#set document(title: "11.1 Distance and Midpoint Formulas; Circles", author: "OpenStax / XYZ Homework")
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== 11.1#h(0.6em)Distance and Midpoint Formulas; Circles

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "20 inches")[$20 #h(0.2em) "inches"$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "x squared minus 18 x plus 81 .")[$x^(2) − 18 x + 81 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x minus 9 squared")[$attach(x − 9, t: 2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by completing the square: #math.equation(block: false, alt: "x squared minus 12 x minus 12 equals 0 .")[$x^(2) − 12 x − 12 = 0 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x equals 6 plus or minus 4 the square root of 3")[$x = 6 ± 4 sqrt(3)$]
]
]

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

#figure(figph[This figure shows two cones placed point to point. They are labeled nappes.], alt: "This figure shows two cones placed point to point. They are labeled nappes.", caption: none)

There are four conics—the #strong[circle], #strong[parabola], #strong[ellipse], and #strong[hyperbola]. The next figure shows how the plane intersecting the double cone results in each curve.

#figure(figph[Each of these four figures shows a double cone intersected by a plane. In the first figure, the plane is perpendicular to the axis of the cones and intersects the bottom cone to form a circle. In the second figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it intersects the base as well. Thus, the curve formed by the intersection is open at both ends. This is labeled parabola. In the third figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it does not intersect the base of the cone. Thus, the curve formed by the intersection is a closed loop, labeled ellipse. In the fourth figure, the plane is parallel to the axis, intersecting both cones. This is labeled hyperbola.], alt: "Each of these four figures shows a double cone intersected by a plane. In the first figure, the plane is perpendicular to the axis of the cones and intersects the bottom cone to form a circle. In the second figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it intersects the base as well. Thus, the curve formed by the intersection is open at both ends. This is labeled parabola. In the third figure, the plane is at an angle to the axis and intersects the bottom cone in such a way that it does not intersect the base of the cone. Thus, the curve formed by the intersection is a closed loop, labeled ellipse. In the fourth figure, the plane is parallel to the axis, intersecting both cones. This is labeled hyperbola.", caption: none)

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.


=== Use the Distance Formula

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

#examplebox("Example 1")[][
Use the rectangular coordinate system to find the distance between the points #math.equation(block: false, alt: "open parenthesis 6 , 4 close parenthesis")[$( 6 , 4 )$] and #math.equation(block: false, alt: "open parenthesis 2 , 1 close parenthesis .")[$( 2 , 1 ) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Plot the two points. Connect the two points #linebreak() with a line. #linebreak() 
Draw a right triangle as if you were going to #linebreak() find slope.], [#figure(figph[A graph on a coordinate plane shows two points, (2,1) and (6,4). A right triangle connects them, illustrating the 'rise' (vertical change) and 'run' (horizontal change) used to find the distance 'd' between the points.], alt: "A graph on a coordinate plane shows two points, (2,1) and (6,4). A right triangle connects them, illustrating the 'rise' (vertical change) and 'run' (horizontal change) used to find the distance 'd' between the points.", caption: none)]),
  [Find the length of each leg.], [#figure(figph[A right triangle has legs of length 3 (rise) and 4 (run), and its hypotenuse is labeled 'd'. This diagram can be used to calculate 'd' using the Pythagorean theorem (3^2 + 4^2 = d^2).], alt: "A right triangle has legs of length 3 (rise) and 4 (run), and its hypotenuse is labeled 'd'. This diagram can be used to calculate 'd' using the Pythagorean theorem (3^2 + 4^2 = d^2).", caption: none)],
  [Use the Pythagorean Theorem to find #emph[d], the #linebreak() distance between the two points.], [#math.equation(block: false, alt: "a squared plus b squared equals c squared")[$a^(2) + b^(2) = c^(2)$]],
  [Substitute in the values.], [#math.equation(block: false, alt: "3 squared plus 4 squared equals d squared")[$3^(2) + 4^(2) = d^(2)$]],
  [Simplify.], [#math.equation(block: false, alt: "9 plus 16 equals d squared")[$#h(0.3em) 9 + 16 = d^(2)$]],
  [], [#math.equation(block: false, alt: "25 equals d squared")[$#h(2em) 25 = d^(2)$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "d equals 5 d equals −5")[$#h(1em) d = 5 #h(2em) cancel(d = −5)$]],
  [Since distance, #emph[d] is positive, we can eliminate #linebreak() #math.equation(block: false, alt: "d equals −5 .")[$d = −5 .$]], [The distance between the points #math.equation(block: false, alt: "open parenthesis 6 , 4 close parenthesis")[$( 6 , 4 )$] and #linebreak() #math.equation(block: false, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$] is 5.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the rectangular coordinate system to find the distance between the points #math.equation(block: false, alt: "open parenthesis 6 , 1 close parenthesis")[$( 6 , 1 )$] and #math.equation(block: false, alt: "open parenthesis 2 , −2 close parenthesis .")[$( 2 , −2 ) .$]

#solutionbox[
#math.equation(block: true, alt: "d equals 5")[$d = 5$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the rectangular coordinate system to find the distance between the points #math.equation(block: false, alt: "open parenthesis 5 , 3 close parenthesis")[$( 5 , 3 )$] and #math.equation(block: false, alt: "open parenthesis −3 , −3 close parenthesis .")[$( −3 , −3 ) .$]

#solutionbox[
#math.equation(block: true, alt: "d equals 10")[$d = 10$]
]
]

#figure(figph[Figure shows a graph with a right triangle. The hypotenuse connects two points, (2, 1) and (6, 4). These are respectively labeled (x1, y1) and (x2, y2). The rise is y2 minus y1, which is 4 minus 1 equals 3. The run is x2 minus x1, which is 6 minus 2 equals 4.], alt: "Figure shows a graph with a right triangle. The hypotenuse connects two points, (2, 1) and (6, 4). These are respectively labeled (x1, y1) and (x2, y2). The rise is y2 minus y1, which is 4 minus 1 equals 3. The run is x2 minus x1, which is 6 minus 2 equals 4.", caption: none)

The method we used in the last example leads us to the formula to find the distance between the two points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis .")[$( x_(2) , y_(2) ) .$]

When we found the length of the horizontal leg we subtracted #math.equation(block: false, alt: "6 minus 2")[$6 − 2$] which is #math.equation(block: false, alt: "x sub 2 minus x sub 1 .")[$x_(2) − x_(1) .$]

When we found the length of the vertical leg we subtracted #math.equation(block: false, alt: "4 minus 1")[$4 − 1$] which is #math.equation(block: false, alt: "y sub 2 minus y sub 1 .")[$y_(2) − y_(1) .$]

If the triangle had been in a different position, we may have subtracted #math.equation(block: false, alt: "x sub 1 minus x sub 2")[$x_(1) − x_(2)$] or #math.equation(block: false, alt: "y sub 1 minus y sub 2 .")[$y_(1) − y_(2) .$] The expressions #math.equation(block: false, alt: "x sub 2 minus x sub 1")[$x_(2) − x_(1)$] and #math.equation(block: false, alt: "x sub 1 minus x sub 2")[$x_(1) − x_(2)$] vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say #math.equation(block: false, alt: "| x sub 2 minus x sub 1 |")[$| x_(2) − x_(1) |$] and #math.equation(block: false, alt: "| y sub 2 minus y sub 1 | .")[$| y_(2) − y_(1) | .$]

In the Pythagorean Theorem, we substitute the general expressions #math.equation(block: false, alt: "| x sub 2 minus x sub 1 |")[$| x_(2) − x_(1) |$] and #math.equation(block: false, alt: "| y sub 2 minus y sub 1 |")[$| y_(2) − y_(1) |$] rather than the numbers.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "a squared plus b squared equals c squared")[$#h(4.7em) a^(2) + b^(2) = c^(2)$]]),
  [Substitute in the values.], [#math.equation(block: false, alt: "open parenthesis | x sub 2 minus x sub 1 | close parenthesis squared plus open parenthesis | y sub 2 minus y sub 1 | close parenthesis squared equals d squared")[$#h(1em) attach(( | x_(2) − x_(1) | ), t: 2) + attach(( | y_(2) − y_(1) | ), t: 2) = d^(2)$]],
  [Squaring the expressions makes them positive, so we eliminate the absolute value bars.], [#math.equation(block: false, alt: "open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared equals d squared")[$#h(1em) attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2) = d^(2)$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "d equals plus or minus the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$#h(4.5em) d = ± sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]],
  [Distance is positive, so eliminate the negative value.], [#math.equation(block: false, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$#h(4.5em) d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]],
))

This is the Distance Formula we use to find the distance #emph[d] between the two points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis .")[$( x_(2) , y_(2) ) .$]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Distance Formula]

The distance #emph[d] between the two points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] is

#math.equation(block: true, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]
]

#examplebox("Example 2")[][
Use the Distance Formula to find the distance between the points #math.equation(block: false, alt: "open parenthesis −5 , −3 close parenthesis")[$( −5 , −3 )$] and #math.equation(block: false, alt: "open parenthesis 7 , 2 close parenthesis .")[$( 7 , 2 ) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the Distance Formula.], [#math.equation(block: false, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]]),
  [Label the points, #math.equation(block: false, alt: "open parenthesis −5 , −3 x sub 1 , y sub 1 close parenthesis , open parenthesis 7 , 2 x sub 2 , y sub 2 close parenthesis")[$( limits(−5 "," −3)^(x_(1) , y_(1)) ) , ( limits(7 "," 2)^(x_(2) , y_(2)) )$] and substitute.], [#math.equation(block: false, alt: "d equals the square root of open parenthesis 7 minus open parenthesis −5 close parenthesis close parenthesis squared plus open parenthesis 2 minus open parenthesis −3 close parenthesis close parenthesis squared")[$d = sqrt(attach(( 7 − ( −5 ) ), t: 2) + attach(( 2 − ( −3 ) ), t: 2))$]],
  [], [],
  [Simplify.], [#math.equation(block: false, alt: "d equals the square root of 12 squared plus 5 squared")[$d = sqrt(12^(2) + 5^(2))$]],
  [], [#math.equation(block: false, alt: "d equals the square root of 144 plus 25")[$d = sqrt(144 + 25)$]],
  [], [#math.equation(block: false, alt: "d equals the square root of 169")[$d = sqrt(169)$]],
  [], [#math.equation(block: false, alt: "d equals 13")[$d = 13$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Distance Formula to find the distance between the points #math.equation(block: false, alt: "open parenthesis −4 , −5 close parenthesis")[$( −4 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 7 close parenthesis .")[$( 5 , 7 ) .$]

#solutionbox[
#math.equation(block: true, alt: "d equals 15")[$d = 15$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Distance Formula to find the distance between the points #math.equation(block: false, alt: "open parenthesis −2 , −5 close parenthesis")[$( −2 , −5 )$] and #math.equation(block: false, alt: "open parenthesis −14 , −10 close parenthesis .")[$( −14 , −10 ) .$]

#solutionbox[
#math.equation(block: true, alt: "d equals 13")[$d = 13$]
]
]

#examplebox("Example 3")[][
Use the Distance Formula to find the distance between the points #math.equation(block: false, alt: "open parenthesis 10 , −4 close parenthesis")[$( 10 , −4 )$] and #math.equation(block: false, alt: "open parenthesis −1 , 5 close parenthesis .")[$( −1 , 5 ) .$] Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the Distance Formula.], [#math.equation(block: false, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]]),
  [Label the points, #math.equation(block: false, alt: "open parenthesis 10 , −4 x sub 1 , y sub 1 close parenthesis , open parenthesis −1 , 5 x sub 2 , y sub 2 close parenthesis")[$( limits(10 "," −4)^(x_(1) , y_(1)) ) , ( limits(−1 "," 5)^(x_(2) , y_(2)) )$] and substitute.], [#math.equation(block: false, alt: "d equals the square root of open parenthesis −1 minus 10 close parenthesis squared plus open parenthesis 5 minus open parenthesis −4 close parenthesis close parenthesis squared")[$d = sqrt(attach(( −1 − 10 ), t: 2) + attach(( 5 − ( −4 ) ), t: 2))$]],
  [Simplify.], [#math.equation(block: false, alt: "d equals the square root of open parenthesis −11 close parenthesis squared plus 9 squared")[$d = sqrt(attach(( −11 ), t: 2) + 9^(2))$]],
  [], [#math.equation(block: false, alt: "d equals the square root of 121 plus 81")[$d = sqrt(121 + 81)$]],
  [], [#math.equation(block: false, alt: "d equals the square root of 202")[$d = sqrt(202)$]],
  [Since 202 is not a perfect square, we can leave the answer in exact form or find a decimal approximation.], [#math.equation(block: false, alt: "d equals the square root of 202; or; d approximately equals 14.2")[$d = sqrt(202) \ #h(1em) "or" \ d ≈ 14.2$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Distance Formula to find the distance between the points #math.equation(block: false, alt: "open parenthesis −4 , −5 close parenthesis")[$( −4 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 3 , 4 close parenthesis .")[$( 3 , 4 ) .$] Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

#solutionbox[
#math.equation(block: true, alt: "d equals the square root of 130 , d approximately equals 11.4")[$d = sqrt(130) , d ≈ 11.4$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Distance Formula to find the distance between the points #math.equation(block: false, alt: "open parenthesis −2 , −5 close parenthesis")[$( −2 , −5 )$] and #math.equation(block: false, alt: "open parenthesis −3 , −4 close parenthesis .")[$( −3 , −4 ) .$] Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

#solutionbox[
#math.equation(block: true, alt: "d equals the square root of 2 , d approximately equals 1.4")[$d = sqrt(2) , d ≈ 1.4$]
]
]


=== Use the Midpoint Formula

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the #emph[x]-coordinates and the average of the #emph[y]-coordinates of the endpoints.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Midpoint Formula]

The midpoint of the line segment whose endpoints are the two points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] is

#math.equation(block: true, alt: "open parenthesis the fraction x sub 1 plus x sub 2 over 2 , the fraction y sub 1 plus y sub 2 over 2 close parenthesis")[$( frac(x_(1) + x_(2), 2) , frac(y_(1) + y_(2), 2) )$]To find the midpoint of a line segment, we find the average of the #emph[x]-coordinates and the average of the #emph[y]-coordinates of the endpoints.
]

#examplebox("Example 4")[][
Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are #math.equation(block: false, alt: "open parenthesis −5 , −4 close parenthesis")[$( −5 , −4 )$] and #math.equation(block: false, alt: "open parenthesis 7 , 2 close parenthesis .")[$( 7 , 2 ) .$] Plot the endpoints and the midpoint on a rectangular coordinate system.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the Midpoint Formula.], [#math.equation(block: false, alt: "open parenthesis the fraction x sub 1 plus x sub 2 over 2 , the fraction y sub 1 plus y sub 2 over 2 close parenthesis")[$( frac(x_(1) + x_(2), 2) , frac(y_(1) + y_(2), 2) )$]]),
  [Label the points, #math.equation(block: false, alt: "open parenthesis −5 , −4 x sub 1 , y sub 1 close parenthesis , open parenthesis 7 , 2 x sub 2 , y sub 2 close parenthesis")[$( limits(−5 "," −4)^(x_(1) , y_(1)) ) , ( limits(7 "," 2)^(x_(2) , y_(2)) )$] #linebreak() and substitute.], [#math.equation(block: false, alt: "open parenthesis the fraction −5 plus 7 over 2 , the fraction −4 plus 2 over 2 close parenthesis")[$( frac(−5 + 7, 2) , frac(−4 + 2, 2) )$]],
  [Simplify.], [#math.equation(block: false, alt: "open parenthesis the fraction 2 over 2 , the fraction −2 over 2 close parenthesis")[$( frac(2, 2) , frac(−2, 2) )$]],
  [], [#math.equation(block: false, alt: "open parenthesis 1 , −1 close parenthesis")[$( 1 , −1 )$] #linebreak() The midpoint of the segment is the point #linebreak() #math.equation(block: false, alt: "open parenthesis 1 , −1 close parenthesis .")[$( 1 , −1 ) .$]],
  [Plot the endpoints and midpoint.], [#figure(figph[A line graph plotted on a coordinate plane, showing a straight line passing through three points: (-5, -4), (1, -1), and (7, 2). The x-axis ranges from -8 to 8, and the y-axis from -8 to 8.], alt: "A line graph plotted on a coordinate plane, showing a straight line passing through three points: (-5, -4), (1, -1), and (7, 2). The x-axis ranges from -8 to 8, and the y-axis from -8 to 8.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are #math.equation(block: false, alt: "open parenthesis −3 , −5 close parenthesis")[$( −3 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 7 close parenthesis .")[$( 5 , 7 ) .$] Plot the endpoints and the midpoint on a rectangular coordinate system.

#solutionbox[
#figure(figph[This graph shows a line segment with endpoints (negative 3, negative 5) and (5, 7) and midpoint (1, negative 1).], alt: "This graph shows a line segment with endpoints (negative 3, negative 5) and (5, 7) and midpoint (1, negative 1).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Midpoint Formula to find the midpoint of the line segment whose endpoints are #math.equation(block: false, alt: "open parenthesis −2 , −5 close parenthesis")[$( −2 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 6 , −1 close parenthesis .")[$( 6 , −1 ) .$] Plot the endpoints and the midpoint on a rectangular coordinate system.

#solutionbox[
#figure(figph[This graph shows a line segment with endpoints (negative 2, negative 5) and (6, negative 1) and midpoint (2, negative 3).], alt: "This graph shows a line segment with endpoints (negative 2, negative 5) and (6, negative 1) and midpoint (2, negative 3).", caption: none)
]
]

Both the Distance Formula and the Midpoint Formula depend on two points, #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis .")[$( x_(2) , y_(2) ) .$] It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

#figure(figph[The distance formula is d equals square root of open parentheses x2 minus x1 close parentheses squared plus open parentheses y2 minus y1 close parentheses squared end of root. This is labeled subtract the coordinates. The midpoint formula is open parentheses open parentheses x1 plus x2 close parentheses upon 2 comma open parentheses y1 plus y2 close parentheses upon 2 close parentheses. This is labeled add the coordinates.], alt: "The distance formula is d equals square root of open parentheses x2 minus x1 close parentheses squared plus open parentheses y2 minus y1 close parentheses squared end of root. This is labeled subtract the coordinates. The midpoint formula is open parentheses open parentheses x1 plus x2 close parentheses upon 2 comma open parentheses y1 plus y2 close parentheses upon 2 close parentheses. This is labeled add the coordinates.", caption: none)


=== Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

#figure(figph[This figure shows a double cone and an intersecting plane, which form a circle.], alt: "This figure shows a double cone and an intersecting plane, which form a circle.", caption: none)

We define a #strong[circle] as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the #emph[center,] #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and the fixed distance is called the #emph[radius], #emph[r], of the circle.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Circle]

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the #strong[center], #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and the fixed distance is called the #strong[radius], #emph[r], of the circle.
]

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We look at a circle in the rectangular coordinate system. #linebreak() The radius is the distance from the center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] to a #linebreak() point on the circle, #math.equation(block: false, alt: "open parenthesis x , y close parenthesis .")[$( x , y ) .$]], [#figure(figph[A circle is plotted on a Cartesian coordinate system with its center at (h, k), a point on the circle at (x, y), and its radius labeled r, connecting the center to the point on the circle.], alt: "A circle is plotted on a Cartesian coordinate system with its center at (h, k), a point on the circle at (x, y), and its radius labeled r, connecting the center to the point on the circle.", caption: none)]),
  [To derive the equation of a circle, we can use the #linebreak() distance formula with the points #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] and the #linebreak() distance, #emph[r].], [#math.equation(block: false, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$#h(0.55em) d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]],
  [Substitute the values.], [#math.equation(block: false, alt: "r equals the square root of open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared")[$#h(0.55em) r = sqrt(attach(( x − h ), t: 2) + attach(( y − k ), t: 2))$]],
  [Square both sides.], [#math.equation(block: false, alt: "r squared equals open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared")[$r^(2) = attach(( x − h ), t: 2) + attach(( y − k ), t: 2)$]],
))

This is the standard form of the equation of a circle with center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and radius, #emph[r].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Standard Form of the Equation a Circle]

The standard form of the equation of a circle with center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and radius, #emph[r], is

#figure(figph[Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.], alt: "Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.", caption: none)
]

#examplebox("Example 5")[][
Write the standard form of the equation of the circle with radius 3 and center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the standard form of the equation of a circle], [#math.equation(block: false, alt: "open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared equals r squared")[$attach(( x − h ), t: 2) + attach(( y − k ), t: 2) = r^(2)$]]),
  [Substitute in the values #math.equation(block: false, alt: "r equals 3 , h equals 0 ,")[$r = 3 , h = 0 ,$] and #math.equation(block: false, alt: "k equals 0 .")[$k = 0 .$]], [#math.equation(block: false, alt: "open parenthesis x minus 0 close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared equals 3 squared")[$attach(( x − 0 ), t: 2) + attach(( y − 0 ), t: 2) = 3^(2)$]],
  [#figure(figph[The image displays the word 'Center:' followed by a stacked coordinate system, showing (h, k) in red above (0, 0) in black, enclosed within a single set of large parentheses. It illustrates the concept of a center point in two forms.], alt: "The image displays the word 'Center:' followed by a stacked coordinate system, showing (h, k) in red above (0, 0) in black, enclosed within a single set of large parentheses. It illustrates the concept of a center point in two forms.", caption: none)], [],
  [Simplify.], [#math.equation(block: false, alt: "x squared plus y squared equals 9")[$x^(2) + y^(2) = 9$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the standard form of the equation of the circle with a radius of 6 and center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]

#solutionbox[
#math.equation(block: true, alt: "x squared plus y squared equals 36")[$x^(2) + y^(2) = 36$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the standard form of the equation of the circle with a radius of 8 and center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]

#solutionbox[
#math.equation(block: true, alt: "x squared plus y squared equals 64")[$x^(2) + y^(2) = 64$]
]
]

In the last example, the center was #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$] Notice what happened to the equation. Whenever the center is #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis ,")[$( 0 , 0 ) ,$] the standard form becomes #math.equation(block: false, alt: "x squared plus y squared equals r squared .")[$x^(2) + y^(2) = r^(2) .$]

#examplebox("Example 6")[][
Write the standard form of the equation of the circle with radius 2 and center #math.equation(block: false, alt: "open parenthesis −1 , 3 close parenthesis .")[$( −1 , 3 ) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the standard form of the equation of a #linebreak() circle.], [#math.equation(block: false, alt: "open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared equals r squared")[$#h(1.3em) attach(( x − h ), t: 2) + attach(( y − k ), t: 2) = r^(2)$]]),
  [Substitute in the values.], [#math.equation(block: false, alt: "open parenthesis x minus open parenthesis −1 close parenthesis close parenthesis squared plus open parenthesis y minus 3 close parenthesis squared equals 2 squared")[$attach(( x − ( −1 ) ), t: 2) + attach(( y − 3 ), t: 2) = 2^(2)$]],
  [#figure(figph[The image displays the word 'Center:' followed by a parenthesis containing two lines of text: the top line reads 'h, k' in red, and the bottom line reads '-1, 3' in black, indicating the center coordinates.], alt: "The image displays the word 'Center:' followed by a parenthesis containing two lines of text: the top line reads 'h, k' in red, and the bottom line reads '-1, 3' in black, indicating the center coordinates.", caption: none)], [],
  [Simplify.], [#math.equation(block: false, alt: "open parenthesis x plus 1 close parenthesis squared plus open parenthesis y minus 3 close parenthesis squared equals 4")[$#h(0.9em) attach(( x + 1 ), t: 2) + attach(( y − 3 ), t: 2) = 4$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the standard form of the equation of the circle with a radius of 7 and center #math.equation(block: false, alt: "open parenthesis 2 , −4 close parenthesis .")[$( 2 , −4 ) .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y plus 4 close parenthesis squared equals 49")[$attach(( x − 2 ), t: 2) + attach(( y + 4 ), t: 2) = 49$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the standard form of the equation of the circle with a radius of 9 and center #math.equation(block: false, alt: "open parenthesis −3 , −5 close parenthesis .")[$( −3 , −5 ) .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis squared plus open parenthesis y plus 5 close parenthesis squared equals 81")[$attach(( x + 3 ), t: 2) + attach(( y + 5 ), t: 2) = 81$]
]
]

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

#examplebox("Example 7")[][
Write the standard form of the equation of the circle with center #math.equation(block: false, alt: "open parenthesis 2 , 4 close parenthesis")[$( 2 , 4 )$] that also contains the point #math.equation(block: false, alt: "open parenthesis −2 , 1 close parenthesis .")[$( −2 , 1 ) .$]

#figure(figph[This graph shows circle with center at (2, 4, radius 5 and a point on the circle minus 2, 1.], alt: "This graph shows circle with center at (2, 4, radius 5 and a point on the circle minus 2, 1.", caption: none)

#solutionbox[
The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center #math.equation(block: false, alt: "open parenthesis 2 , 4 close parenthesis")[$( 2 , 4 )$] and point #math.equation(block: false, alt: "open parenthesis −2 , 1 close parenthesis")[$( −2 , 1 )$]

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the Distance Formula to find the radius.], [#math.equation(block: false, alt: "r equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$#h(4em) r = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]]),
  [Substitute the values. #math.equation(block: false, alt: "open parenthesis 2 , 4 x sub 1 , y sub 1 close parenthesis , open parenthesis −2 , 1 x sub 2 , y sub 2 close parenthesis")[$( limits(2 "," 4)^(x_(1) , y_(1)) ) , ( limits(−2 "," 1)^(x_(2) , y_(2)) )$]], [#math.equation(block: false, alt: "r equals the square root of open parenthesis −2 minus 2 close parenthesis squared plus open parenthesis 1 minus 4 close parenthesis squared")[$#h(4em) r = sqrt(attach(( −2 − 2 ), t: 2) + attach(( 1 − 4 ), t: 2))$]],
  [Simplify.], [#math.equation(block: false, alt: "r equals the square root of open parenthesis −4 close parenthesis squared plus open parenthesis −3 close parenthesis squared")[$#h(4em) r = sqrt(attach(( −4 ), t: 2) + attach(( −3 ), t: 2))$]],
  [], [#math.equation(block: false, alt: "r equals the square root of 16 plus 9")[$#h(4em) r = sqrt(16 + 9)$]],
  [], [#math.equation(block: false, alt: "r equals the square root of 25")[$#h(4em) r = sqrt(25)$]],
  [], [#math.equation(block: false, alt: "r equals 5")[$#h(4em) r = 5$]],
))

Now that we know the radius, #math.equation(block: false, alt: "r equals 5 ,")[$r = 5 ,$] and the center, #math.equation(block: false, alt: "open parenthesis 2 , 4 close parenthesis ,")[$( 2 , 4 ) ,$] we can use the standard form of the equation of a circle to find the equation.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the standard form of the equation of a circle.], [#math.equation(block: false, alt: "open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared equals r squared")[$#h(2em) attach(( x − h ), t: 2) + attach(( y − k ), t: 2) = r^(2)$]]),
  [Substitute in the values.], [#math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y minus 4 close parenthesis squared equals 5 squared")[$#h(2em) attach(( x − 2 ), t: 2) + attach(( y − 4 ), t: 2) = 5^(2)$]],
  [Simplify.], [#math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y minus 4 close parenthesis squared equals 25")[$#h(2em) attach(( x − 2 ), t: 2) + attach(( y − 4 ), t: 2) = 25$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the standard form of the equation of the circle with center #math.equation(block: false, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$] that also contains the point #math.equation(block: false, alt: "open parenthesis −2 , −2 close parenthesis .")[$( −2 , −2 ) .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 25")[$attach(( x − 2 ), t: 2) + attach(( y − 1 ), t: 2) = 25$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the standard form of the equation of the circle with center #math.equation(block: false, alt: "open parenthesis 7 , 1 close parenthesis")[$( 7 , 1 )$] that also contains the point #math.equation(block: false, alt: "open parenthesis −1 , −5 close parenthesis .")[$( −1 , −5 ) .$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 7 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 100")[$attach(( x − 7 ), t: 2) + attach(( y − 1 ), t: 2) = 100$]
]
]


=== Graph a Circle

Any equation of the form #math.equation(block: false, alt: "open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared equals r squared")[$attach(( x − h ), t: 2) + attach(( y − k ), t: 2) = r^(2)$] is the standard form of the equation of a #strong[circle] with center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and radius, #emph[r.] We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from #emph[x] and #emph[y]. In the next example, the equation has #math.equation(block: false, alt: "x plus 2 ,")[$x + 2 ,$] so we need to rewrite the addition as subtraction of a negative.

#examplebox("Example 8")[][
Find the center and radius, then graph the circle: #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 9 .")[$attach(( x + 2 ), t: 2) + attach(( y − 1 ), t: 2) = 9 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The equation of a circle is displayed as (x + 2)^2 + (y - 1)^2 = 9.], alt: "The equation of a circle is displayed as (x + 2)^2 + (y - 1)^2 = 9.", caption: none)]),
  [Use the standard form of the equation of a circle. #linebreak() Identify the center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$] and radius, #emph[r].], [#figure(figph[The standard form of a circle's equation, (x-h)^2 + (y-k)^2 = r^2, is shown with blue arrows indicating how to identify the values for h, k, and r in an example equation: (x - (-2))^2 + (y - 1)^2 = 3^2.], alt: "The standard form of a circle's equation, (x-h)^2 + (y-k)^2 = r^2, is shown with blue arrows indicating how to identify the values for h, k, and r in an example equation: (x - (-2))^2 + (y - 1)^2 = 3^2.", caption: none)],
  [], [Center: #math.equation(block: false, alt: "open parenthesis −2 , 1 close parenthesis")[$( −2 , 1 )$] radius: 3],
  [Graph the circle.], [#figure(figph[A circle with center at (-2, 1) and radius r=3 plotted on a Cartesian plane.], alt: "A circle with center at (-2, 1) and radius r=3 plotted on a Cartesian plane.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared plus open parenthesis y plus 4 close parenthesis squared equals 4 .")[$attach(( x − 3 ), t: 2) + attach(( y + 4 ), t: 2) = 4 .$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 3 , −4 close parenthesis")[$( 3 , −4 )$] with a radius of 2. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows a circle with center at (3, negative 4) and a radius of 2.], alt: "This graph shows a circle with center at (3, negative 4) and a radius of 2.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 16 .")[$attach(( x − 3 ), t: 2) + attach(( y − 1 ), t: 2) = 16 .$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] with a radius of 4. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (3, 1) and a radius of 4.], alt: "This graph shows circle with center at (3, 1) and a radius of 4.", caption: none)
]
]

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of #math.equation(block: false, alt: "x squared , y squared")[$x^(2) , y^(2)$] to be one.

#examplebox("Example 9")[][
Find the center and radius and then graph the circle, #math.equation(block: false, alt: "4 x squared plus 4 y squared equals 64 .")[$4 x^(2) + 4 y^(2) = 64 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed reading '4x^2 + 4y^2 = 64' in black text against a white background.], alt: "A mathematical equation is displayed reading '4x^2 + 4y^2 = 64' in black text against a white background.", caption: none)]),
  [Divide each side by 4.], [#figure(figph[The equation x^2 + y^2 = 16 is displayed on a white background, representing a circle centered at the origin with a radius of 4.], alt: "The equation x^2 + y^2 = 16 is displayed on a white background, representing a circle centered at the origin with a radius of 4.", caption: none)],
  [Use the standard form of the equation of a circle. #linebreak() Identify the center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$] and radius, #emph[r].], [#figure(figph[Substitution of h=0, k=0, and r=4 into the standard equation of a circle, (x-h)^2 + (y-k)^2 = r^2, resulting in (x-0)^2 + (y-0)^2 = 4^2.], alt: "Substitution of h=0, k=0, and r=4 into the standard equation of a circle, (x-h)^2 + (y-k)^2 = r^2, resulting in (x-0)^2 + (y-0)^2 = 4^2.", caption: none)],
  [], [Center: #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] radius: 4],
  [Graph the circle.], [#figure(figph[A circle centered at the origin (0,0) with a radius of 4, displayed on a Cartesian coordinate system.], alt: "A circle centered at the origin (0,0) with a radius of 4, displayed on a Cartesian coordinate system.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "3 x squared plus 3 y squared equals 27")[$3 x^(2) + 3 y^(2) = 27$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] with a radius of 3. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (0, 0) and a radius of 3.], alt: "This graph shows circle with center at (0, 0) and a radius of 3.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "5 x squared plus 5 y squared equals 125")[$5 x^(2) + 5 y^(2) = 125$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] with a radius of 5. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (0, 0) and a radius of 5.], alt: "This graph shows circle with center at (0, 0) and a radius of 5.", caption: none)
]
]

If we expand the equation from , #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 9 ,")[$attach(( x + 2 ), t: 2) + attach(( y − 1 ), t: 2) = 9 ,$] the equation of the circle looks very different.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 9")[$attach(( x + 2 ), t: 2) + attach(( y − 1 ), t: 2) = 9$]]),
  [Square the binomials.], [#math.equation(block: false, alt: "x squared plus 4 x plus 4 plus y squared minus 2 y plus 1 equals 9")[$x^(2) + 4 x + 4 + y^(2) − 2 y + 1 = 9$]],
  [Arrange the terms in descending degree order, and get zero on the right], [#math.equation(block: false, alt: "x squared plus y squared plus 4 x minus 2 y minus 4 equals 0")[$x^(2) + y^(2) + 4 x − 2 y − 4 = 0$]],
))

This form of the equation is called the general form of the equation of the #strong[circle].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[General Form of the Equation of a Circle]

The general form of the equation of a circle is

#math.equation(block: true, alt: "x squared plus y squared plus a x plus b y plus c equals 0")[$x^(2) + y^(2) + a x + b y + c = 0$]
]

If we are given an equation in general form, we can change it to standard form by completing the squares in both #emph[x] and #emph[y]. Then we can graph the circle using its center and radius.

#examplebox("Example 10")[][
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "x squared plus y squared minus 4 x minus 6 y plus 4 equals 0 .")[$x^(2) + y^(2) − 4 x − 6 y + 4 = 0 .$]

#solutionbox[
We need to rewrite this general form into standard form in order to find the center and radius. #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The image shows the equation of a circle in general form: x^2 + y^2 - 4x - 6y + 4 = 0.], alt: "The image shows the equation of a circle in general form: x^2 + y^2 - 4x - 6y + 4 = 0.", caption: none)]),
  [Group the #emph[x]-terms and #emph[y]-terms. #linebreak() Collect the constants on the right side.], [#figure(figph[The image displays the algebraic equation x^2 - 4x + y^2 - 6y = -4 on a white background, representing a circle in its general form.], alt: "The image displays the algebraic equation x^2 - 4x + y^2 - 6y = -4 on a white background, representing a circle in its general form.", caption: none)],
  [Complete the squares.], [#figure(figph[The equation x² - 4x + 4 + y² - 6y + 9 = -4 + 4 + 9, a step in completing the square for a circle's equation.], alt: "The equation x² - 4x + 4 + y² - 6y + 9 = -4 + 4 + 9, a step in completing the square for a circle's equation.", caption: none)],
  [Rewrite as binomial squares.], [#figure(figph[The equation of a circle, (x-2)^2 + (y-3)^2 = 9, is displayed against a white background.], alt: "The equation of a circle, (x-2)^2 + (y-3)^2 = 9, is displayed against a white background.", caption: none)],
  [Identify the center and radius.], [Center: #math.equation(block: false, alt: "open parenthesis 2 , 3 close parenthesis")[$( 2 , 3 )$] radius: 3],
  [Graph the circle.], [#figure(figph[Graphic representation of a circle in a coordinate plane with center at (2, 3) and radius of 3 units.], alt: "Graphic representation of a circle in a coordinate plane with center at (2, 3) and radius of 3 units.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "x squared plus y squared minus 6 x minus 8 y plus 9 equals 0 .")[$x^(2) + y^(2) − 6 x − 8 y + 9 = 0 .$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 3 , 4 close parenthesis")[$( "3" , "4" )$] with a radius of 4. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (3, 4) and a radius of 4.], alt: "This graph shows circle with center at (3, 4) and a radius of 4.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "x squared plus y squared plus 6 x minus 2 y plus 1 equals 0 .")[$x^(2) + y^(2) + 6 x − 2 y + 1 = 0 .$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis − 3 , 1 close parenthesis")[$( "−" "3" , "1" )$] with a radius of 3. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (negative 3, 1) and a radius of 3.], alt: "This graph shows circle with center at (negative 3, 1) and a radius of 3.", caption: none)
]
]

In the next example, there is a #emph[y]-term and a #math.equation(block: false, alt: "y squared")[$y^(2)$]-term. But notice that there is no #emph[x]-term, only an #math.equation(block: false, alt: "x squared")[$x^(2)$]-term. We have seen this before and know that it means #emph[h] is 0. We will need to complete the square for the #emph[y] terms, but not for the #emph[x] terms.

#examplebox("Example 11")[][
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "x squared plus y squared plus 8 y equals 0 .")[$x^(2) + y^(2) + 8 y = 0 .$]

#solutionbox[
We need to rewrite this general form into standard form in order to find the center and radius. #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The equation x^2 + y^2 + 8y = 0 is shown in black text against a white background, representing a circle in coordinate geometry.], alt: "The equation x^2 + y^2 + 8y = 0 is shown in black text against a white background, representing a circle in coordinate geometry.", caption: none)]),
  [Group the #emph[x]-terms and #emph[y]-terms.], [#figure(figph[A mathematical equation is displayed on a white background, reading x^2 + y^2 + 8y = 0. The equation represents a circle in standard form.], alt: "A mathematical equation is displayed on a white background, reading x^2 + y^2 + 8y = 0. The equation represents a circle in standard form.", caption: none)],
  [There are no constants to collect on the #linebreak() right side.], [],
  [Complete the square for #math.equation(block: false, alt: "y squared plus 8 y .")[$y^(2) + 8 y .$]], [#figure(figph[The equation x^2 + y^2 + 8y + 16 = 0 + 16 is displayed, with the number 16 highlighted in red on both sides, indicating a step in an algebraic manipulation, likely completing the square.], alt: "The equation x^2 + y^2 + 8y + 16 = 0 + 16 is displayed, with the number 16 highlighted in red on both sides, indicating a step in an algebraic manipulation, likely completing the square.", caption: none)],
  [Rewrite as binomial squares.], [#figure(figph[A mathematical equation representing a circle is shown, specifically (x-0)^2 + (y+4)^2 = 16. This is the standard form equation for a circle centered at (0, -4) with a radius of 4.], alt: "A mathematical equation representing a circle is shown, specifically (x-0)^2 + (y+4)^2 = 16. This is the standard form equation for a circle centered at (0, -4) with a radius of 4.", caption: none)],
  [Identify the center and radius.], [Center: #math.equation(block: false, alt: "open parenthesis 0 , −4 close parenthesis")[$( 0 , −4 )$] radius: 4],
  [Graph the circle.], [#figure(figph[A circle is plotted on a coordinate plane, centered at (0, -4) with a radius of 4. The x-axis extends from -6 to 6, and the y-axis from -10 to 2, with grid lines every unit.], alt: "A circle is plotted on a coordinate plane, centered at (0, -4) with a radius of 4. The x-axis extends from -6 to 6, and the y-axis from -10 to 2, with grid lines every unit.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "x squared plus y squared minus 2 x minus 3 equals 0 .")[$x^(2) + y^(2) − 2 x − 3 = 0 .$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] with a radius of 2. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (1, 0) and a radius of 2.], alt: "This graph shows circle with center at (1, 0) and a radius of 2.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Find the center and radius, then ⓑ graph the circle: #math.equation(block: false, alt: "x squared plus y squared minus 12 y plus 11 equals 0 .")[$x^(2) + y^(2) − 12 y + 11 = 0 .$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 0 , 6 close parenthesis")[$( 0 , 6 )$] with a radius of 5. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (0, 6) and a radius of 5.], alt: "This graph shows circle with center at (0, 6) and a radius of 5.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

- #link("https://openstax.org/l/37distmidcircle")[Distance-Midpoint Formulas and Circles]
- #link("https://openstax.org/l/37distmid2pts")[Finding the Distance and Midpoint Between Two Points]
- #link("https://openstax.org/l/37stformcircle")[Completing the Square to Write Equation in Standard Form of a Circle]
]


=== Key Concepts

- #strong[Distance Formula:] The distance #emph[d] between the two points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] is #linebreak() 
#math.equation(block: true, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared")[$d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2))$]
- #strong[Midpoint Formula:] The midpoint of the line segment whose endpoints are the two points #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] is #linebreak() 
#math.equation(block: true, alt: "open parenthesis the fraction x sub 1 plus x sub 2 over 2 , the fraction y sub 1 plus y sub 2 over 2 close parenthesis")[$( frac(x_(1) + x_(2), 2) , frac(y_(1) + y_(2), 2) )$] #linebreak() 
To find the midpoint of a line segment, we find the average of the #emph[x]-coordinates and the average of the #emph[y]-coordinates of the endpoints.
- #strong[Circle:] A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the #emph[center,] #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and the fixed distance is called the #emph[radius, r,] of the circle.
- #strong[Standard Form of the Equation a Circle:] The standard form of the equation of a circle with center, #math.equation(block: false, alt: "open parenthesis h , k close parenthesis ,")[$( h , k ) ,$] and radius, #emph[r,] is #linebreak() 
#figure(figph[Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.], alt: "Figure shows circle with center at (h, k) and a radius of r. A point on the circle is labeled x, y. The formula is open parentheses x minus h close parentheses squared plus open parentheses y minus k close parentheses squared equals r squared.", caption: none)
- #strong[General Form of the Equation of a Circle:] The general form of the equation of a circle is #linebreak() 
#math.equation(block: true, alt: "x squared plus y squared plus a x plus b y plus c equals 0")[$x^(2) + y^(2) + a x + b y + c = 0$]


==== Practice Makes Perfect

#strong[Use the Distance Formula]

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

#math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 4 close parenthesis")[$( 5 , 4 )$]

#solutionbox[
#math.equation(block: true, alt: "d equals 5")[$d = 5$]
]

#math.equation(block: false, alt: "open parenthesis −4 , −3 close parenthesis")[$( −4 , −3 )$] and #math.equation(block: false, alt: "open parenthesis 2 , 5 close parenthesis")[$( 2 , 5 )$]

#math.equation(block: false, alt: "open parenthesis −4 , −3 close parenthesis")[$( −4 , −3 )$] and #math.equation(block: false, alt: "open parenthesis 8 , 2 close parenthesis")[$( 8 , 2 )$]

#solutionbox[
13
]

#math.equation(block: false, alt: "open parenthesis −7 , −3 close parenthesis")[$( −7 , −3 )$] and #math.equation(block: false, alt: "open parenthesis 8 , 5 close parenthesis")[$( 8 , 5 )$]

#math.equation(block: false, alt: "open parenthesis −1 , 4 close parenthesis")[$( −1 , 4 )$] and #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$]

#solutionbox[
5
]

#math.equation(block: false, alt: "open parenthesis −1 , 3 close parenthesis")[$( −1 , 3 )$] and #math.equation(block: false, alt: "open parenthesis 5 , −5 close parenthesis")[$( 5 , −5 )$]

#math.equation(block: false, alt: "open parenthesis 1 , −4 close parenthesis")[$( 1 , −4 )$] and #math.equation(block: false, alt: "open parenthesis 6 , 8 close parenthesis")[$( 6 , 8 )$]

#solutionbox[
13
]

#math.equation(block: false, alt: "open parenthesis −8 , −2 close parenthesis")[$( −8 , −2 )$] and #math.equation(block: false, alt: "open parenthesis 7 , 6 close parenthesis")[$( 7 , 6 )$]

#math.equation(block: false, alt: "open parenthesis −3 , −5 close parenthesis")[$( −3 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]

#solutionbox[
#math.equation(block: true, alt: "d equals 3 the square root of 5 , d approximately equals 6.7")[$d = 3 sqrt(5) , d ≈ 6.7$]
]

#math.equation(block: false, alt: "open parenthesis −1 , −2 close parenthesis")[$( −1 , −2 )$] and #math.equation(block: false, alt: "open parenthesis −3 , 4 close parenthesis")[$( −3 , 4 )$]

#math.equation(block: false, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$] and #math.equation(block: false, alt: "open parenthesis 1 , 7 close parenthesis")[$( 1 , 7 )$]

#solutionbox[
#math.equation(block: true, alt: "d equals 2 the square root of 17 , d approximately equals 8.2")[$d = 2 sqrt(17) , d ≈ 8.2$]
]

#math.equation(block: false, alt: "open parenthesis −4 , −5 close parenthesis")[$( −4 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 7 , 4 close parenthesis")[$( 7 , 4 )$]

#strong[Use the Midpoint Formula]

In the following exercises, ⓐ find the midpoint of the line segment whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

#math.equation(block: false, alt: "open parenthesis 0 , −5 close parenthesis")[$( 0 , −5 )$] and #math.equation(block: false, alt: "open parenthesis 4 , −3 close parenthesis")[$( 4 , −3 )$]

#solutionbox[
ⓐ Midpoint: #math.equation(block: false, alt: "open parenthesis 2 , −4 close parenthesis")[$( 2 , −4 )$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows line segment with endpoints (0, negative 5) and (4, negative 3) and midpoint (2, negative 4).], alt: "This graph shows line segment with endpoints (0, negative 5) and (4, negative 3) and midpoint (2, negative 4).", caption: none)
]

#math.equation(block: false, alt: "open parenthesis −2 , −6 close parenthesis")[$( −2 , −6 )$] and #math.equation(block: false, alt: "open parenthesis 6 , −2 close parenthesis")[$( 6 , −2 )$]

#math.equation(block: false, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$] and #math.equation(block: false, alt: "open parenthesis 4 , −2 close parenthesis")[$( 4 , −2 )$]

#solutionbox[
ⓐ Midpoint: #math.equation(block: false, alt: "open parenthesis 3 the fraction 1 over 2 , −1 the fraction 1 over 2 close parenthesis")[$( 3 frac(1, 2) , −1 frac(1, 2) )$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows line segment with endpoints (3, negative 1) and (4, negative 2) and midpoint (3 and a half, negative 1 and a half).], alt: "This graph shows line segment with endpoints (3, negative 1) and (4, negative 2) and midpoint (3 and a half, negative 1 and a half).", caption: none)
]

#math.equation(block: false, alt: "open parenthesis −3 , −3 close parenthesis")[$( −3 , −3 )$] and #math.equation(block: false, alt: "open parenthesis 6 , −1 close parenthesis")[$( 6 , −1 )$]

#strong[Write the Equation of a Circle in Standard Form]

In the following exercises, write the standard form of the equation of the circle with the given radius and center #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis .")[$( 0 , 0 ) .$]

Radius: 7

#solutionbox[
#math.equation(block: true, alt: "x squared plus y squared equals 49")[$x^(2) + y^(2) = 49$]
]

Radius: 9

Radius: #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]

#solutionbox[
#math.equation(block: true, alt: "x squared plus y squared equals 2")[$x^(2) + y^(2) = 2$]
]

Radius: #math.equation(block: false, alt: "the square root of 5")[$sqrt(5)$]

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Radius: 1, center: #math.equation(block: false, alt: "open parenthesis 3 , 5 close parenthesis")[$( 3 , 5 )$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis squared plus open parenthesis y minus 5 close parenthesis squared equals 1")[$attach(( x − 3 ), t: 2) + attach(( y − 5 ), t: 2) = 1$]
]

Radius: 10, center: #math.equation(block: false, alt: "open parenthesis −2 , 6 close parenthesis")[$( −2 , 6 )$]

Radius: #math.equation(block: false, alt: "2.5 ,")[$2.5 ,$] center: #math.equation(block: false, alt: "open parenthesis 1.5 , −3.5 close parenthesis")[$( 1.5 , −3.5 )$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 1.5 close parenthesis squared plus open parenthesis y plus 3.5 close parenthesis squared equals 6.25")[$attach(( x − 1.5 ), t: 2) + attach(( y + 3.5 ), t: 2) = 6.25$]
]

Radius: #math.equation(block: false, alt: "1.5 ,")[$1.5 ,$] center: #math.equation(block: false, alt: "open parenthesis −5.5 , −6.5 close parenthesis")[$( −5.5 , −6.5 )$]

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Center #math.equation(block: false, alt: "open parenthesis 3 , −2 close parenthesis")[$( 3 , −2 )$] with point #math.equation(block: false, alt: "open parenthesis 3 , 6 close parenthesis")[$( 3 , 6 )$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis squared plus open parenthesis y plus 2 close parenthesis squared equals 64")[$attach(( x − 3 ), t: 2) + attach(( y + 2 ), t: 2) = 64$]
]

Center #math.equation(block: false, alt: "open parenthesis 6 , −6 close parenthesis")[$( 6 , −6 )$] with point #math.equation(block: false, alt: "open parenthesis 2 , −3 close parenthesis")[$( 2 , −3 )$]

Center #math.equation(block: false, alt: "open parenthesis 4 , 4 close parenthesis")[$( 4 , 4 )$] with point #math.equation(block: false, alt: "open parenthesis 2 , 2 close parenthesis")[$( 2 , 2 )$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 4 close parenthesis squared plus open parenthesis y minus 4 close parenthesis squared equals 8")[$attach(( x − 4 ), t: 2) + attach(( y − 4 ), t: 2) = 8$]
]

Center #math.equation(block: false, alt: "open parenthesis −5 , 6 close parenthesis")[$( −5 , 6 )$] with point #math.equation(block: false, alt: "open parenthesis −2 , 3 close parenthesis")[$( −2 , 3 )$]

#strong[Graph a Circle]

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

#math.equation(block: true, alt: "open parenthesis x plus 5 close parenthesis squared plus open parenthesis y plus 3 close parenthesis squared equals 1")[$attach(( x + 5 ), t: 2) + attach(( y + 3 ), t: 2) = 1$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis −5 , −3 close parenthesis")[$( −5 , −3 )$] with a radius of 1. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows a circle with center at (negative 5, negative 3) and a radius of 1.], alt: "This graph shows a circle with center at (negative 5, negative 3) and a radius of 1.", caption: none)
]

#math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y minus 3 close parenthesis squared equals 9")[$attach(( x − 2 ), t: 2) + attach(( y − 3 ), t: 2) = 9$]

#math.equation(block: true, alt: "open parenthesis x minus 4 close parenthesis squared plus open parenthesis y plus 2 close parenthesis squared equals 16")[$attach(( x − 4 ), t: 2) + attach(( y + 2 ), t: 2) = 16$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 4 , −2 close parenthesis")[$( 4 , −2 )$] with a radius of 4. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (4, negative 2) and a radius of 4.], alt: "This graph shows circle with center at (4, negative 2) and a radius of 4.", caption: none)
]

#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis squared plus open parenthesis y minus 5 close parenthesis squared equals 4")[$attach(( x + 2 ), t: 2) + attach(( y − 5 ), t: 2) = 4$]

#math.equation(block: true, alt: "x squared plus open parenthesis y plus 2 close parenthesis squared equals 25")[$x^(2) + attach(( y + 2 ), t: 2) = 25$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 0 , −2 close parenthesis")[$( 0 , −2 )$] with a radius of 5. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (negative 2, 5) and a radius of 5.], alt: "This graph shows circle with center at (negative 2, 5) and a radius of 5.", caption: none)
]

#math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis squared plus y squared equals 36")[$attach(( x − 1 ), t: 2) + y^(2) = 36$]

#math.equation(block: true, alt: "open parenthesis x minus 1.5 close parenthesis squared plus open parenthesis y plus 2.5 close parenthesis squared equals 0.25")[$attach(( x − 1.5 ), t: 2) + attach(( y + 2.5 ), t: 2) = 0.25$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 1.5 , −2.5 close parenthesis")[$( 1.5 , −2.5 )$] with a radius of #math.equation(block: false, alt: "0.5 .")[$0.5 .$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (1.5, 2.5) and a radius of 0.5], alt: "This graph shows circle with center at (1.5, 2.5) and a radius of 0.5", caption: none)
]

#math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis squared plus open parenthesis y minus 3 close parenthesis squared equals the fraction 9 over 4")[$attach(( x − 1 ), t: 2) + attach(( y − 3 ), t: 2) = frac(9, 4)$]

#math.equation(block: true, alt: "x squared plus y squared equals 64")[$x^(2) + y^(2) = 64$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] with a radius of 8. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (0, 0) and a radius of 8.], alt: "This graph shows circle with center at (0, 0) and a radius of 8.", caption: none)
]

#math.equation(block: true, alt: "x squared plus y squared equals 49")[$x^(2) + y^(2) = 49$]

#math.equation(block: true, alt: "2 x squared plus 2 y squared equals 8")[$2 x^(2) + 2 y^(2) = 8$]

#solutionbox[
ⓐ The circle is centered at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] with a radius of 2. #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (0, 0) and a radius of 2.], alt: "This graph shows circle with center at (0, 0) and a radius of 2.", caption: none)
]

#math.equation(block: true, alt: "6 x squared plus 6 y squared equals 216")[$6 x^(2) + 6 y^(2) = 216$]

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

#math.equation(block: true, alt: "x squared plus y squared plus 2 x plus 6 y plus 9 equals 0")[$x^(2) + y^(2) + 2 x + 6 y + 9 = 0$]

#solutionbox[
ⓐ Center: #math.equation(block: false, alt: "open parenthesis −1 , −3 close parenthesis ,")[$( −1 , −3 ) ,$] radius: 1 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (negative 1, negative 3) and a radius of 1.], alt: "This graph shows circle with center at (negative 1, negative 3) and a radius of 1.", caption: none)
]

#math.equation(block: true, alt: "x squared plus y squared minus 6 x minus 8 y equals 0")[$x^(2) + y^(2) − 6 x − 8 y = 0$]

#math.equation(block: true, alt: "x squared plus y squared minus 4 x plus 10 y minus 7 equals 0")[$x^(2) + y^(2) − 4 x + 10 y − 7 = 0$]

#solutionbox[
ⓐ Center: #math.equation(block: false, alt: "open parenthesis 2 , −5 close parenthesis ,")[$( 2 , −5 ) ,$] radius: 6 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (2, negative 5) and a radius of 6.], alt: "This graph shows circle with center at (2, negative 5) and a radius of 6.", caption: none)
]

#math.equation(block: true, alt: "x squared plus y squared plus 12 x minus 14 y plus 21 equals 0")[$x^(2) + y^(2) + 12 x − 14 y + 21 = 0$]

#math.equation(block: true, alt: "x squared plus y squared plus 6 y plus 5 equals 0")[$x^(2) + y^(2) + 6 y + 5 = 0$]

#solutionbox[
ⓐ Center: #math.equation(block: false, alt: "open parenthesis 0 , −3 close parenthesis ,")[$( 0 , −3 ) ,$] radius: 2 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (0, negative 3) and a radius of 2.], alt: "This graph shows circle with center at (0, negative 3) and a radius of 2.", caption: none)
]

#math.equation(block: true, alt: "x squared plus y squared minus 10 y equals 0")[$x^(2) + y^(2) − 10 y = 0$]

#math.equation(block: true, alt: "x squared plus y squared plus 4 x equals 0")[$x^(2) + y^(2) + 4 x = 0$]

#solutionbox[
ⓐ Center: #math.equation(block: false, alt: "open parenthesis −2 , 0 close parenthesis ,")[$( −2 , 0 ) ,$] radius: 2  #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This graph shows circle with center at (negative 2, 0) and a radius of 2.], alt: "This graph shows circle with center at (negative 2, 0) and a radius of 2.", caption: none)
]

#math.equation(block: true, alt: "x squared plus y squared minus 14 x plus 13 equals 0")[$x^(2) + y^(2) − 14 x + 13 = 0$]


==== Writing Exercises

Explain the relationship between the distance formula and the equation of a circle.

#solutionbox[
Answers will vary.
]

Is a circle a function? Explain why or why not.

In your own words, state the definition of a circle.

#solutionbox[
Answers will vary.
]

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[A self-assessment table for math skills, where students rate their confidence in using distance and midpoint formulas, and writing or graphing circle equations. Options: Confidently, With some help, No-I don't get it!], alt: "A self-assessment table for math skills, where students rate their confidence in using distance and midpoint formulas, and writing or graphing circle equations. Options: Confidently, With some help, No-I don't get it!", caption: none)

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.