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== 1.3#h(0.6em)Circles


    
      
=== The Distance Formula


      Delbert is hiking in the Santa Monica mountains, and he would like to know the distance from the Sycamore Canyon trail head, located at 12-C on his map, to the Coyote Trail junction, located at 8-F, as shown below.


      #figure(figph[map of mountains], alt: "map of mountains", caption: none)


      Each interval on the map represents one kilometer. Delbert remembers the #strong[Pythagorean theorem], and uses the map coordinates to label the sides of a right triangle. The distance he wants is the hypotenuse of the triangle, so


      #math.equation(block: true, alt: "d squared, equals 4 squared plus 3 squared equals 16 plus 9 equals 25; d, equals the square root of 25 equals 5")[$d^(2) & = 4^(2) + 3^(2) = 16 + 9 = 25 \ d & = sqrt(25) = 5$]
      The straight-line distance to Coyote junction is about 5 kilometers.


      The formula for the distance between two points is obtained in the same way.


      We first label a right triangle with points #math.equation(block: false, alt: "P sub 1")[$P_(1)$] and #math.equation(block: false, alt: "P sub 2")[$P_(2)$] on opposite ends of the hypotenuse. (See the figure at right.) The sides of the triangle have lengths #math.equation(block: false, alt: "vertical bar x sub 2 minus x sub 1 vertical bar")[$| x_(2) − x_(1) |$] and #math.equation(block: false, alt: "vertical bar y sub 2 minus y sub 1 vertical bar")[$| y_(2) − y_(1) |$].  We can use the Pythagorean theorem to calculate the distance between #math.equation(block: false, alt: "P sub 1")[$P_(1)$] and #math.equation(block: false, alt: "P sub 2")[$P_(2)$]:


      #math.equation(block: true, alt: "d squared equals 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")[$d^(2) = ( x_(2) − x_(1) )^(2) + ( y_(2) − y_(1) )^(2)$]
      #figure(figph[distance formula], alt: "distance formula", caption: none)


      Taking the (positive) square root of each side of this equation gives us the #strong[distance formula].


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


        The distance #math.equation(block: false, alt: "d")[$d$] between two points #math.equation(block: false, alt: "P sub 1 open parenthesis x sub 1 , y sub 1 close parenthesis")[$P_(1) ( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "P sub 2 open parenthesis x sub 2 , y sub 2 close parenthesis")[$P_(2) ( x_(2) , y_(2) )$] 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")[$bold(italic(d)) = sqrt(( bold(italic(x))_(2) − bold(italic(x))_(1) )^(2) + ( bold(italic(y))_(2) − bold(italic(y))_(1) )^(2))$]
]


      #examplebox("Example 1")[][
Find the distance between #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 3 close parenthesis")[$( 4 , 3 )$].


        
        #solutionbox[
We substitute #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$] for #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 4 , 3 close parenthesis")[$( 4 , 3 )$] for #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$] in the distance formula to obtain


          #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; equals the square root of open parenthesis 4 minus 2 close parenthesis squared plus open bracket 3 minus open parenthesis minus 1 close parenthesis close bracket squared; equals the square root of 4 plus 16 equals the square root of 20 equals 2 the square root of 5")[$d & = sqrt(( x_(2) − x_(1) )^(2) + ( y_(2) − y_(1) )^(2)) \  & = sqrt(( 4 − 2 )^(2) + [ 3 − ( − 1 ) ]^(2)) \  & = sqrt(4 + 16) = sqrt(20) = 2 sqrt(5)$]
          #figure(figph[distance between two points], alt: "distance between two points", caption: none)


          The exact value of the distance, shown at right, is #math.equation(block: false, alt: "2 the square root of 5")[$2 sqrt(5)$] units.  We obtain the same answer if we use #math.equation(block: false, alt: "open parenthesis 4 , 3 close parenthesis")[$( 4 , 3 )$] for #math.equation(block: false, alt: "P sub 1")[$P_(1)$] and #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$] for #math.equation(block: false, alt: "P sub 2")[$P_(2)$]:


          #math.equation(block: true, alt: "d, equals the square root of open parenthesis 2 minus 4 close parenthesis squared plus open bracket open parenthesis minus 1 close parenthesis minus 3 close bracket squared; equals the square root of 4 plus 16 equals the square root of 20 equals 2 the square root of 5")[$d & = sqrt(( 2 − 4 )^(2) + [ ( − 1 ) − 3 ]^(2)) \  & = sqrt(4 + 16) = sqrt(20) = 2 sqrt(5)$]
          We can use a calculator to obtain an approximation for this value, and find


          #math.equation(block: true, alt: "2 the square root of 5 approximately equals 2 open parenthesis 2.236 close parenthesis equals 4.472")[$2 sqrt(5) ≈ 2 ( 2.236 ) = 4.472$]
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In the previous example, the radical #math.equation(block: false, alt: "the square root of 4 plus 16")[$sqrt(4 + 16)$] cannot be simplified to #math.equation(block: false, alt: "the square root of 4 plus the square root of 16")[$sqrt(4) + sqrt(16)$]. (Do you remember why not?)
]


      
        + Find the distance between the points #math.equation(block: false, alt: "open parenthesis minus 5 , 3 close parenthesis")[$( − 5 , 3 )$] and #math.equation(block: false, alt: "open parenthesis 3 , minus 9 close parenthesis")[$( 3 , − 9 )$].
+ Plot the points on a Cartesian grid and show how the Pythagorean theorem is used to calculate the distance.


        
          + #math.equation(block: false, alt: "4 the square root of 13")[$4 sqrt(13)$]
+ The points #math.equation(block: false, alt: "open parenthesis minus 5 , 3 close parenthesis")[$( − 5 , 3 )$] and #math.equation(block: false, alt: "open parenthesis 3 , minus 9 close parenthesis")[$( 3 , − 9 )$] are two vertices of a right triangle with one horizontal leg of length 8 and one vertical leg of length 12. The distance between #math.equation(block: false, alt: "open parenthesis minus 5 , 3 close parenthesis")[$( − 5 , 3 )$] and #math.equation(block: false, alt: "open parenthesis 3 , minus 9 close parenthesis")[$( 3 , − 9 )$] is the hypotenuse #math.equation(block: false, alt: "c")[$c$].

#math.equation(block: true, alt: "c squared, equals 8 squared plus 12 squared; equals 208")[$c^(2) & = 8^(2) + 12^(2) \  & = 208$]so #math.equation(block: false, alt: "c equals the square root of 208 equals 4 the square root of 13")[$" " c = sqrt(208) = 4 sqrt(13)$]

#figure(figph[distance formula], alt: "distance formula", caption: none)


        
      
    
    
      
=== Equation for a Circle


      A #strong[circle] is the set of all points in a plane that lie at a given distance, called the #strong[radius], from a fixed point called the #strong[center].  We can use the distance formula to find an equation for a circle.


      The circle shown below has its center at the origin, #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], and its radius is #math.equation(block: false, alt: "r")[$r$].


      Now, the distance from the origin to any point #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis")[$P ( x , y )$] on the circle is #math.equation(block: false, alt: "r")[$r$].  Therefore,


      #math.equation(block: true, alt: "the square root of open parenthesis x minus 0 close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared equals r")[$sqrt(( x − 0 )^(2) + ( y − 0 )^(2)) = r$]
      or, squaring both sides,


      #math.equation(block: true, alt: "open parenthesis x minus 0 close parenthesis squared plus open parenthesis y minus 0 close parenthesis squared equals r squared")[$( x − 0 )^(2) + ( y − 0 )^(2) = r^(2)$]
      #figure(figph[circle centered at origin], alt: "circle centered at origin", caption: none)


      Because every point on the circle must satisfy this equation, we have found an equation for the circle.


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


        The equation for a #strong[circle] of radius #math.equation(block: false, alt: "r")[$r$] centered at the origin is


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


      #examplebox("Example 2")[][
Find two points on the circle #math.equation(block: false, alt: "x squared plus y squared equals 4")[$" " x^(2) + y^(2) = 4 " "$] with #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "minus 1")[$− 1$].


        
        #solutionbox[
We substitute #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$] into the equation for the circle, and solve for #math.equation(block: false, alt: "y")[$y$].


          #math.equation(block: true, alt: "open parenthesis minus 1 squared close parenthesis plus y squared, equals 4; y squared, equals 4 minus 1 equals 3; y, equals plus or minus the square root of 3")[$( attach(− 1, t: 2) ) + y^(2) & = 4 \ y^(2) & = 4 − 1 = 3 \ y & = ± sqrt(3)$]
          The points are #math.equation(block: false, alt: "open parenthesis minus 1 , the square root of 3 close parenthesis")[$( − 1 , sqrt(3) )$] and #math.equation(block: false, alt: "open parenthesis minus 1 , the square root of 3 close parenthesis")[$( − 1 , sqrt(3) )$], as shown at right. Note that #math.equation(block: false, alt: "the square root of 3 approximately equals 1.732")[$sqrt(3) ≈ 1.732$].


          #figure(figph[circle with radius=2], alt: "circle with radius=2", caption: none)
]
]


      
        Find the coordinates of two points on the circle #math.equation(block: false, alt: "x squared plus y squared equals 1")[$" " " " x^(2) + y^(2) = 1 " " " "$] with #math.equation(block: false, alt: "y")[$y$]-coordinate #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$].


        
          Substitute #math.equation(block: false, alt: "y equals the fraction 1 over 2")[$y = display(frac(1, 2))$] into the equation and solve for #math.equation(block: false, alt: "x")[$x$] to find #math.equation(block: false, alt: "x squared equals the fraction 3 over 4")[$" " x^(2) = display(frac(3, 4)) " "$] so #math.equation(block: false, alt: "x equals the fraction the square root of 3 over 2")[$" " x = display(frac(sqrt(3), 2)) " "$] or #math.equation(block: false, alt: "x equals the fraction minus the square root of 3 over 2")[$" " x = display(frac(− sqrt(3), 2))$]. The points are #math.equation(block: false, alt: "open parenthesis the fraction the square root of 3 over 2 , the fraction 1 over 2 close parenthesis")[$" " ( display(frac(sqrt(3), 2)) , display(frac(1, 2)) )$], #math.equation(block: false, alt: "open parenthesis the fraction minus the square root of 3 over 2 , the fraction 1 over 2 close parenthesis")[$" " ( display(frac(− sqrt(3), 2)) , display(frac(1, 2)) )$]


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


        The circle in the exercise above, #math.equation(block: false, alt: "x squared plus y squared equals 1 ,")[$" " " " bold(italic(x))^(2) + bold(italic(y))^(2) = 1 , " " " "$] which is centered at the origin and has radius 1 unit, is called the #strong[unit circle].
]


    
    
      
=== Rational and Irrational Numbers


      Every common fraction, such as #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$], can be written in many equivalent forms, including a decimal form.  For example,


      #math.equation(block: true, alt: "the fraction 3 over 4 equals the fraction 6 over 8 equals the fraction 75 over 100 equals 0.75 and the fraction 5 over 11 equals the fraction 20 over 44 equals the fraction 50 over 110 equals 0.45 bar")[$frac(3, 4) = frac(6, 8) = frac(75, 100) = 0.75 " " " " " " " " " " "and" " " " " " " " " " " frac(5, 11) = frac(20, 44) = frac(50, 110) = 0. limits(45)^(―)$]
      where #math.equation(block: false, alt: "0.45 bar")[$0. limits(45)^(―)$] is the repeating decimal #math.equation(block: false, alt: "0.45454545. . . .")[$" " 0.45454545 . . . .$]


      Because the fraction bar denotes division, a fraction is a quotient of two integers, and we can calculate its decimal form by dividing the denominator into the numerator.


      #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Rational Number]


        Any number (including fractions) that can be written as a quotient of two integers #math.equation(block: false, alt: "the fraction a over b , where b not equal to 0 ,")[$" " " " display(frac(a, b)) , " " " " "where" " " " " b ≠ 0 , " " " "$] is called a #strong[rational number].
]


      The decimal form of a rational number is either a #strong[terminating decimal], such as #math.equation(block: false, alt: "0.75")[$0.75$], or a #strong[repeating decimal], such as #math.equation(block: false, alt: "0.45 bar")[$0. limits(45)^(―)$]. Thus, we can always write down an exact decimal equivalent for a rational number, although we may choose to round off a particularly long or unwieldy decimal. For example,


      #math.equation(block: true, alt: "the fraction 3 over 7 equals 0.428571 bar approximately equals 0.43")[$frac(3, 7) = 0. limits(428571)^(―) ≈ 0.43$]
      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Note that #math.equation(block: false, alt: "0.43")[$0.43$] is #emph[not] exactly equal to #math.equation(block: false, alt: "the fraction 3 over 7")[$display(frac(3, 7))$]; it is an #emph[approximation] for #math.equation(block: false, alt: "the fraction 3 over 7")[$display(frac(3, 7))$], just as #math.equation(block: false, alt: "0.33")[$0.33$] is an approximation for #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$]. In our work, it will be important to distinguish between exact values and approximations.
]


      #notebox("Definition", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Irrational Number]


        An #strong[irrational number] is one that #emph[cannot] be written as a quotient of two integers #math.equation(block: false, alt: "the fraction a over b")[$display(frac(a, b))$],where #math.equation(block: false, alt: "b not equal to 0")[$b ≠ 0$].
]


      Examples of irrational numbers are #math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$], #math.equation(block: false, alt: "the cube root of 586")[$root(3, 586)$], and #math.equation(block: false, alt: "π")[$π$]. The decimal form of an irrational number is nonterminating and nonrepeating. The first few digits of the examples mentioned are


      #math.equation(block: true, alt: "the square root of 3, equals 1.732050807568. . .; the cube root of 586, equals 8.368209391204. . .; π, equals 3.141592653589. . .")[$sqrt(3) & = 1.732050807568 . . . \ root(3, 586) & = 8.368209391204 . . . \ π & = 3.141592653589 . . .$]
      but none of these decimal forms ever ends. Thus, #emph[we cannot write down an exact decimal equivalent for an irrational number]. The best we can do is give a decimal approximation, no matter how many digits we include.


      #examplebox("Example 3")[][
Which values are exact, and which are approximations?


          + #math.equation(block: false, alt: "the fraction 7 over 32 → 0.21875")[$display(frac(7, 32)) → 0.21875$]
+ #math.equation(block: false, alt: "the square root of 8 → 2.828427125")[$sqrt(8) → 2.828427125$]
+ #math.equation(block: false, alt: "the square root of 0.16 → 0.4")[$sqrt(0.16) → 0.4$]
+ #math.equation(block: false, alt: "the fraction π over 2 → 1.570796327")[$display(frac(π, 2)) → 1.570796327$]


        
        #solutionbox[
+ Because #math.equation(block: false, alt: "the fraction 7 over 32")[$display(frac(7, 32))$] is a rational number, it has an exact decimal equivalent.  Divide 7 by 32 to see that #math.equation(block: false, alt: "the fraction 7 over 32 equals 0.21875")[$display(frac(7, 32)) = 0.21875$].
+ Because 8 is not a perfect square, #math.equation(block: false, alt: "the square root of 8")[$sqrt(8)$] is irrational, so #math.equation(block: false, alt: "2.828427125")[$2.828427125$] is not the exact value of #math.equation(block: false, alt: "the square root of 8")[$sqrt(8)$].
+ #math.equation(block: false, alt: "the square root of 0.16 equals the square root of the fraction 16 over 100 equals the fraction 4 over 10 equals 0.4")[$sqrt(0.16) = sqrt(display(frac(16, 100))) = display(frac(4, 10)) = 0.4$], so this value is exact.
+ Because #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] is an irrational number, it has no decimal equivalent, so #math.equation(block: false, alt: "1.570796327")[$1.570796327$] is an approximation.
]
]


      
        For which numbers can you give an exact decimal equivalent?


        + #math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]
+ #math.equation(block: false, alt: "the fraction the square root of 16 over 3")[$display(frac(sqrt(16), 3))$]
+ #math.equation(block: false, alt: "2 π")[$2 π$]
+ #math.equation(block: false, alt: "the fraction 25 over 17")[$display(frac(25, 17))$]


        
          (b) and (d) are rational numbers.


        
      
    
    
      
=== Circumference and Area


      Recall from geometry that the circumference of a circle is proportional to its radius.


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


        The #strong[circumference] of a circle of radius #math.equation(block: false, alt: "r")[$r$] is given by


        #math.equation(block: true, alt: "C equals 2 π r")[$bold(italic(C)) = 2 bold(italic(π)) bold(italic(r))$]
]


      The number #math.equation(block: false, alt: "π")[$π$] gives the ratio of the circumference of any circle to its diameter.  It is an irrational number, #math.equation(block: false, alt: "π approximately equals 3.14159")[$π ≈ 3.14159$].


      The length of a portion, or arc, of a circle, is called its #strong[arclength].


      #examplebox("Example 4")[][
Delbert baked an apple pie with diameter 8 inches. If Francine cuts herself a 60° wedge, what is the arclength of the curved edge?


        
        #solutionbox[
The radius of the pie is 4 inches, so its circumference is #math.equation(block: false, alt: "2 π open parenthesis 4 close parenthesis")[$2 π ( 4 )$] inches.


          A 60° wedge, shown at right, is #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] of the entire pie, so its edge is #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] of the circumference. The exact length of the arc is thus


          #math.equation(block: true, alt: "the fraction 1 over 6 open parenthesis 8 π close parenthesis equals the fraction 4 π over 3 inches.")[$display(frac(1, 6)) ( 8 π ) = display(frac(4 π, 3)) " inches."$]
          #figure(figph[circle with a wedge], alt: "circle with a wedge", caption: none)


          Using a calculator, we find that #math.equation(block: false, alt: "the fraction 4 π over 3 equals 4.19")[$display(frac(4 π, 3)) = 4.19$] rounded to two decimal places, so the length of the curved edge is between 4 and 4#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] inches.
]
]


      
        What is the arclength of the curved edge of a 60° wedge cut from a blueberry pie of diameter 10 inches?


        
          #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] of the circumference is 
#math.equation(block: false, alt: "the fraction 1 over 6 open parenthesis 10 π close parenthesis equals the fraction 5 π over 3")[$" " display(frac(1, 6)) ( 10 π ) = display(frac(5 π, 3)) " "$] inches or approximately 5.24 inches.


        
      
      The area of a circle is proportional to the square of its radius.


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


        The #strong[area] of a circle of radius #math.equation(block: false, alt: "r")[$r$] is given by


        #math.equation(block: true, alt: "A equals π r squared")[$bold(italic(A)) = bold(italic(π)) bold(italic(r))^(2)$]
]


      A portion of a circle shaped like a pie-shaped wedge is called a #strong[sector].


      #examplebox("Example 5")[][
What is the area of Francine's slice of apple pie in the previous example?


        
        #solutionbox[
As we saw in the previous example, Francine's sector of the pie is #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] of the entire pie, so its area #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] of the area of the whole pie, or


          #math.equation(block: true, alt: "the fraction 1 over 6 π open parenthesis 4 squared close parenthesis equals the fraction 8 π over 3 square inches")[$display(frac(1, 6)) π ( 4^(2) ) = display(frac(8 π, 3)) " " " " "square inches"$]
          The area of the wedge is #math.equation(block: false, alt: "the fraction 8 π over 3")[$display(frac(8 π, 3))$], or about 8.38 square inches.
]
]


      
        What is the area of a 60° wedge cut from a blueberry pie of diameter 10 inches?


        
          #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] of the area is #math.equation(block: false, alt: "the fraction 25 π over 6 approximately equals 13.09 square inches")[$" " display(frac(25 π, 6)) ≈ 13.09 " " " " "square inches"$]


        
      
      Review the following skills you will need for this section.


      #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
True of False.


        + #math.equation(block: false, alt: "the square root of a squared plus b squared equals a plus b")[$sqrt(a^(2) + b^(2)) = a + b$]
+ #math.equation(block: false, alt: "the square root of 36 plus 64 equals 6 plus 8")[$sqrt(36 + 64) = 6 + 8$]
+ #math.equation(block: false, alt: "the square root of 16 x to the power 4 equals 4 x squared")[$sqrt(16 x^(4)) = 4 x^(2)$]
+ #math.equation(block: false, alt: "the square root of 2 x the square root of 3 y equals the square root of 6 x y")[$sqrt(2 x) sqrt(3 y) = sqrt(6 x y)$]
+ #math.equation(block: false, alt: "the square root of 5 x plus the square root of 3 x equals the square root of 8 x")[$sqrt(5 x) + sqrt(3 x) = sqrt(8 x)$]
+ #math.equation(block: false, alt: "the square root of 4 plus N equals 2 plus the square root of N")[$sqrt(4 + N) = 2 + sqrt(N)$]
+ #math.equation(block: false, alt: "the square root of the fraction x over 4 equals the fraction the square root of x over 2")[$sqrt(display(frac(x, 4))) = display(frac(sqrt(x), 2))$]
+ #math.equation(block: false, alt: "the square root of the fraction 3 over 2 equals the fraction the square root of 6 over 2")[$sqrt(display(frac(3, 2))) = display(frac(sqrt(6), 2))$]


        Algebra Refresher Answers


        + False
+ False
+ True
+ True
+ False
+ False
+ True
+ True
]


    
    
      
=== Section 1.3 Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Rational number
- Irrational number
- Repeating decimal
- Circle
- Radius
- Unit circle
- Circumference
- Arclength


      
      
        
==== Concepts


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

The distance #math.equation(block: false, alt: "d")[$d$] between two points #math.equation(block: false, alt: "P sub 1 open parenthesis x sub 1 , y sub 1 close parenthesis")[$P_(1) ( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "P sub 2 open parenthesis x sub 2 , y sub 2 close parenthesis")[$P_(2) ( x_(2) , y_(2) )$] 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")[$d = sqrt(( x_(2) − x_(1) )^(2) + ( y_(2) − y_(1) )^(2))$]
]
+ Any number that can be written as a quotient of two integers #math.equation(block: false, alt: "the fraction a over b , where b not equal to 0 ,")[$" " " " display(frac(a, b)) , " " " " "where" " " " " b ≠ 0 ,$] is called a #strong[rational number].  The decimal form of a rational number is either a #strong[terminating decimal] or a #strong[repeating decimal].
+ An #strong[irrational number] is one that #emph[cannot] be written as a quotient of two integers #math.equation(block: false, alt: "the fraction a over b , where b not equal to 0")[$" " " " display(frac(a, b)) , " " " " "where" " " " " b ≠ 0$].  We cannot write down an exact decimal equivalent for an irrational number.
+ A #strong[circle] is the set of all points in a plane that lie at a given distance, called the #strong[radius], from a fixed point called the #strong[center].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Circle]

The equation for a #strong[circle] of radius #math.equation(block: false, alt: "r")[$r$] centered at the origin is

#math.equation(block: true, alt: "x squared plus y squared equals r squared")[$x^(2) + y^(2) = r^(2)$]
]
+ The circle #math.equation(block: false, alt: "x squared plus y squared equals 1")[$" " x^(2) + y^(2) = 1 " "$], which is centered at the origin and has radius 1 unit, is called the #strong[unit circle].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Circumference of a Circle]

The #strong[circumference] of a circle of radius #math.equation(block: false, alt: "r")[$r$] is given by

#math.equation(block: true, alt: "C equals 2 π r")[$C = 2 π r$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Area of a Circle]

The #strong[area] of a circle of radius #math.equation(block: false, alt: "r")[$r$] is given by

#math.equation(block: true, alt: "A equals π r squared")[$A = π r^(2)$]
]


      
      
        
==== Study Questions


        + Explain why 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")[$d = sqrt(( x_(2) − x_(1) )^(2) + ( y_(2) − y_(1) )^(2))$],  cannot be simplified to #math.equation(block: false, alt: "open parenthesis x sub 2 minus x sub 1 close parenthesis plus open parenthesis y sub 2 minus y sub 1 close parenthesis")[$( x_(2) − x_(1) ) + ( y_(2) − y_(1) )$].
+ What is a unit circle, and what is its equation?
+ Can you give an exact decimal value for #math.equation(block: false, alt: "π")[$π$]?
+ What is the arclength of a semicircle of radius #math.equation(block: false, alt: "r")[$r$]?


      
      
        
==== Skills


        Practice each skill in the Homework Problems listed.


        + Find the distance between two points    \#1-18
+ Distinguish between exact values and approximations    \#19-24
+ Graph a circle    \#25-30, 35-40
+ Find and use the equation for a circle    \#31-34, 39-42, 55-56
+ Find the length of a fraction of a circle    \#35-38, 43-54
+ Find the area of a sector of a circle    \#43-50


      
    
    
      
=== Homework 1.3


      
        Leanne is sailing and is currently 3 miles west and 5 miles south of the harbor.  She heads directly towards an island that is 8 miles west and 7 miles north of the harbor.  How far is Leanne from the island?


        
          13 miles


        
      
      
        Dominic is 100 meters east and 250 meters north of Kristen.  He is walking directly towards a tree that is 220 meters east and 90 meters north of Kristen.  How far is Dominic from the tree?


      
      For Problems 3–6, find the distance between the points.  Give your answer as an exact value, then as a decimal rounded to hundredths.


      
        #figure(figph[two points], alt: "two points", caption: none)


        
          10, 10.00


        
      
      
        #figure(figph[two points], alt: "two points", caption: none)


      
      
        #figure(figph[two points], alt: "two points", caption: none)


        
          #math.equation(block: true, alt: "4 the square root of 5 approximately equals 8.94")[$4 sqrt(5) ≈ 8.94$]


        
      
      
        #figure(figph[two points], alt: "two points", caption: none)


      
      For Problems 7–12, find the distance between the points.


      
        #math.equation(block: true, alt: "open parenthesis 1 , 1 close parenthesis , open parenthesis 4 , 5 close parenthesis")[$" " ( 1 , 1 ) , " " ( 4 , 5 )$]


        
          5


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 1 , 1 close parenthesis , open parenthesis 5 , 9 close parenthesis")[$" " ( − 1 , 1 ) , " " ( 5 , 9 )$]


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


        
          #math.equation(block: true, alt: "2 the square root of 5")[$2 sqrt(5)$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 5 , minus 4 close parenthesis , open parenthesis minus 1 , 1 close parenthesis")[$" " ( 5 , − 4 ) , " " ( − 1 , 1 )$]


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


        
          5


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


      
      
        Sketch a triangle with vertices #math.equation(block: false, alt: "open parenthesis 10 , 1 close parenthesis , open parenthesis 3 , 1 close parenthesis , open parenthesis 5 , 9 close parenthesis")[$( 10 , 1 ) , " " ( 3 , 1 ) , " " ( 5 , 9 )$] and find its perimeter.  Round your answer to tenths.


        
          #figure(figph[triangle on grid], alt: "triangle on grid", caption: none)


          #math.equation(block: true, alt: "24.7")[$" " " " 24.7$]


        
      
      
        Sketch a triangle with vertices #math.equation(block: false, alt: "open parenthesis minus 1 , 5 close parenthesis , open parenthesis 8 , minus 7 close parenthesis , open parenthesis 4 , 1 close parenthesis")[$( − 1 , 5 ) , " " ( 8 , − 7 ) , " " ( 4 , 1 )$] and find its perimeter. Round your answer to tenths.


      
      
        + Write an expression for the distance between the points #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] and #math.equation(block: false, alt: "open parenthesis minus 3 , 4 close parenthesis")[$( − 3 , 4 )$].
+ Write an equation that says "the distance between the points #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] and #math.equation(block: false, alt: "open parenthesis minus 3 , 4 close parenthesis")[$( − 3 , 4 )$] is 5 units."


        
          + #math.equation(block: false, alt: "the square root of open parenthesis x plus 3 close parenthesis squared plus open parenthesis y minus 4 close parenthesis squared")[$sqrt(( x + 3 )^(2) + ( y − 4 )^(2))$]
+ #math.equation(block: false, alt: "the square root of open parenthesis x plus 3 close parenthesis squared plus open parenthesis y minus 4 close parenthesis squared equals 5")[$sqrt(( x + 3 )^(2) + ( y − 4 )^(2)) = 5$]


        
      
      
        + Write an expression for the distance between the points #math.equation(block: false, alt: "open parenthesis minus 6 , minus 1 close parenthesis")[$( − 6 , − 1 )$] and #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$].
+ Write an equation that says "the point #math.equation(block: false, alt: "open parenthesis minus 6 , minus 1 close parenthesis")[$( − 6 , − 1 )$] is 3 units from the point #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]."


      
      Interpret the equations in Problems 17 and 18 as statements about distance.


      
        #math.equation(block: true, alt: "the square root of open parenthesis x minus 4 close parenthesis squared plus open parenthesis y plus 1 close parenthesis squared equals 3")[$sqrt(( x − 4 )^(2) + ( y + 1 )^(2)) = 3$]


        
          The distance between the points #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] and #math.equation(block: false, alt: "open parenthesis 4 , minus 1 close parenthesis")[$( 4 , − 1 )$] is 3 units.


        
      
      
        #math.equation(block: true, alt: "the square root of open parenthesis minus 2 minus h close parenthesis squared plus open parenthesis 5 minus k close parenthesis squared equals l")[$sqrt(( − 2 − h )^(2) + ( 5 − k )^(2)) = l$]


      
      For Problems 19–22,


      + give an exact answer,
+ round your answer to hundredths.


      
        How long is the diagonal of a square whose side is 6 centimeters?


        
          + #math.equation(block: false, alt: "6 the square root of 2")[$6 sqrt(2) " "$]cm
+ 8.49 cm


        
      
      
        How long is the side of a cube whose volume is 80 cubic feet?


      
      
        What is the area of a circle whose radius is 5 inches?


        
          + #math.equation(block: false, alt: "25 π")[$25 π " "$]sq in
+ 78.54 sq in


        
      
      
        What is the circumference of a circle whose radius is 5 meters?


      
      In Problems 23 and 24, decide whether the decimal form is an exact value or an approximation.


      
        + #math.equation(block: false, alt: "the fraction the square root of 3 over 2 → 0.866025")[$display(frac(sqrt(3), 2)) → 0.866025$]
+ #math.equation(block: false, alt: "the fraction the square root of 16 over 3 → 1.333333")[$display(frac(sqrt(16), 3)) → 1.333333$]
+ #math.equation(block: false, alt: "the fraction 3 π over 2 → 4.712390")[$display(frac(3 π, 2)) → 4.712390$]
+ #math.equation(block: false, alt: "the fraction 5 over 7 → 0.714285 bar")[$display(frac(5, 7)) → 0. limits(714285)^(―)$]


        
          + approximation
+ approximation
+ approximation
+ exact


        
      
      
        + #math.equation(block: false, alt: "the fraction π over 6 → 0.52536")[$display(frac(π, 6)) → 0.52536$]
+ #math.equation(block: false, alt: "the fraction 5 over 32 → 0.15625")[$display(frac(5, 32)) → 0.15625$]
+ #math.equation(block: false, alt: "the fraction the square root of 196 over 32 → 0.4375")[$display(frac(sqrt(196), 32)) → 0.4375$]
+ #math.equation(block: false, alt: "the fraction the square root of 8 over 4 → 0.707107")[$display(frac(sqrt(8), 4)) → 0.707107$]


      
      
        + Complete the table of values for the equation #math.equation(block: false, alt: "x squared plus y squared equals 25.")[$x^(2) + y^(2) = 25.$]#figure(table(
  columns: 12, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$],
))
+ Plot the points and graph the equation.

#figure(figph[8x8 grid], alt: "8x8 grid", caption: none)


        
          + #figure(table(
  columns: 12, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "plus or minus 3")[$± 3$]], [#math.equation(block: false, alt: "plus or minus 4")[$± 4$]], [#math.equation(block: false, alt: "plus or minus the square root of 21")[$± sqrt(21)$]], [#math.equation(block: false, alt: "plus or minus 2 the square root of 6")[$± 2 sqrt(6)$]], [#math.equation(block: false, alt: "plus or minus 5")[$± 5$]], [#math.equation(block: false, alt: "plus or minus 2 the square root of 6")[$± 2 sqrt(6)$]], [#math.equation(block: false, alt: "plus or minus the square root of 21")[$± sqrt(21)$]], [#math.equation(block: false, alt: "plus or minus 4")[$± 4$]], [#math.equation(block: false, alt: "plus or minus 3")[$± 3$]], [#math.equation(block: false, alt: "0")[$0$]],
))
+ #figure(figph[circle on grid], alt: "circle on grid", caption: none)


        
      
      
        + Complete the table of values for the equation #math.equation(block: false, alt: "x squared plus y squared equals 1.")[$x^(2) + y^(2) = 1.$]#figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 0.8")[$− 0.8$]], [#math.equation(block: false, alt: "minus 0.6")[$− 0.6$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], [#math.equation(block: false, alt: "1")[$1$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$],
))
+ Plot the points and graph the equation.

#figure(figph[0.2 grid], alt: "0.2 grid", caption: none)


      
      
        + Sketch a graph of all points in the plane that lie 6 units from the origin.
+ Write an equation for your graph.


        
          + #figure(figph[circle on grid], alt: "circle on grid", caption: none)
+ #math.equation(block: false, alt: "x squared plus y squared equals 36")[$x^(2) + y^(2) = 36$]


        
      
      
        + Sketch a graph of all points in the plane that lie #math.equation(block: false, alt: "the square root of 12")[$sqrt(12)$] units from the origin.
+ Write an equation for your graph.


      
      
        + Sketch a graph of all points in the plane that lie less than 3 units from the origin.
+ Write an inequality for your graph.


        
          + #figure(figph[circle on grid], alt: "circle on grid", caption: none)
+ #math.equation(block: false, alt: "x squared plus y squared less than 9")[$x^(2) + y^(2) < 9$]


        
      
      
        + Sketch a graph of all points in the plane that lie more than 5 units from the origin.
+ Write an inequality for your graph.


      
      
        + Explain why the solutions of the equation #math.equation(block: false, alt: "x squared plus y squared equals 16")[$x^(2) + y^(2) = 16$] must have #math.equation(block: false, alt: "minus 4 less than or equal to x less than or equal to 4")[$− 4 ≤ x ≤ 4$].
+ What does part (a) tell you about the graph of the equation?


        
          + No real value of #math.equation(block: false, alt: "y")[$y$] can satisfy #math.equation(block: false, alt: "x squared plus y squared equals 16")[$x^(2) + y^(2) = 16$] unless #math.equation(block: false, alt: "minus 4 less than or equal to x less than or equal to 4.")[$− 4 ≤ x ≤ 4.$]
+ The graph has no points where #math.equation(block: false, alt: "x greater than 4")[$x > 4$] and no points where #math.equation(block: false, alt: "x less than minus 4.")[$x < − 4.$]


        
      
      
        + Explain why the solutions of the equation #math.equation(block: false, alt: "x squared plus y squared equals 100")[$x^(2) + y^(2) = 100$] must have #math.equation(block: false, alt: "minus 10 less than or equal to x less than or equal to 10.")[$− 10 ≤ x ≤ 10.$]
+ What does part (a) tell you about the graph of the equation?


      
      
        The point #math.equation(block: false, alt: "open parenthesis 1 , minus 3 close parenthesis")[$( 1 , − 3 )$] lies on a circle centered at the origin.  What is the radius of the circle?


        
          #math.equation(block: true, alt: "the square root of 10")[$sqrt(10)$]


        
      
      
        The point #math.equation(block: false, alt: "open parenthesis minus 2 , the square root of 6 close parenthesis")[$( − 2 , sqrt(6) )$] lies on a circle centered at the origin.  What is the radius of the circle?


      
      For Problems 35–38,


      + Graph each equation.
+ Find the circumference of the circle.


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


        #figure(figph[8x8 grid], alt: "8x8 grid", caption: none)


        
          + #figure(figph[circle], alt: "circle", caption: none)
+ #math.equation(block: false, alt: "12 π")[$12 π$]


        
      
      
        #math.equation(block: true, alt: "x squared plus y squared equals 16")[$x^(2) + y^(2) = 16$]


        #figure(figph[8x8 grid], alt: "8x8 grid", caption: none)


      
      
        #math.equation(block: true, alt: "4 x squared plus 4 y squared equals 16")[$4 x^(2) + 4 y^(2) = 16$]


        #figure(figph[8x8 grid], alt: "8x8 grid", caption: none)


        
          + #figure(figph[circle], alt: "circle", caption: none)
+ #math.equation(block: false, alt: "4 π")[$4 π$]


        
      
      
        #math.equation(block: true, alt: "2 x squared plus 2 y squared equals 18")[$2 x^(2) + 2 y^(2) = 18$]


        #figure(figph[8x8 grid], alt: "8x8 grid", caption: none)


      
      
        Give the coordinates of two points on the circle in Problem 35 that have #math.equation(block: false, alt: "y equals minus 4")[$y = − 4$]. Plot those points on your graph.


        
          #math.equation(block: true, alt: "open parenthesis minus 2 the square root of 5 , minus 4 close parenthesis , open parenthesis 2 the square root of 5 , minus 4 close parenthesis")[$( − 2 sqrt(5) , − 4 ) , ( 2 sqrt(5) , − 4 )$]


          #figure(figph[circle], alt: "circle", caption: none)


        
      
      
        Give the coordinates of two points on the circle in Problem 35 that have #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$]. Plot those points on your graph.


      
      For Problems 41 and 42, find the coordinates of the points on the unit circle.


      
        #figure(figph[circle], alt: "circle", caption: none)


        
          #math.equation(block: true, alt: "P open parenthesis the fraction 1 over 2 , the fraction the square root of 3 over 2 close parenthesis , Q open parenthesis the fraction 1 over 2 , the fraction minus the square root of 3 over 2 close parenthesis , R open parenthesis the fraction minus 3 over 4 , the fraction the square root of 7 over 4 close parenthesis , S open parenthesis the fraction minus 3 over 4 , the fraction minus the square root of 7 over 4 close parenthesis")[$P ( display(frac(1, 2)) , display(frac(sqrt(3), 2)) ) , Q ( display(frac(1, 2)) , display(frac(− sqrt(3), 2)) ) , R ( display(frac(− 3, 4)) , display(frac(sqrt(7), 4)) ) , S ( display(frac(− 3, 4)) , display(frac(− sqrt(7), 4)) )$]


        
      
      
        #figure(figph[circle], alt: "circle", caption: none)


      
      
        A circular herb garden has diameter 40 feet, and is divided into 8 equal sectors.


        + What is the central angle of each sector?
+ What is the length of the circular edge of each sector?
+ What is the area of each sector?


        #figure(figph[circle], alt: "circle", caption: none)


        
          + #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]
+ #math.equation(block: false, alt: "5 π")[$5 π$] ft
+ #math.equation(block: false, alt: "50 π")[$50 π$] sq ft


        
      
      
        A dart board is 18 inches in diameter, divided into 20 sectors of equal size. (For this exercise, we will ignore the bulls-eye and the fact that  the sectors are further subdivided.)


        + What is the central angle of each sector?
+ What is the length of the circular edge of each sector?
+ What is the area of each sector?


        #figure(figph[circle], alt: "circle", caption: none)


      
      For Problems 45–50,


      + What fraction of one revolution is the central angle?
+ What is the area of the shaded sector?
+ What is the length of the shaded arc?


      
        #figure(figph[circle], alt: "circle", caption: none)


        
          + #math.equation(block: false, alt: "the fraction 2 over 5")[$display(frac(2, 5))$]
+ #math.equation(block: false, alt: "40 π")[$40 π$] sq ft
+ #math.equation(block: false, alt: "8 π")[$8 π$] ft


        
      
      
        #figure(figph[circle], alt: "circle", caption: none)


      
      
        #figure(figph[circle], alt: "circle", caption: none)


        
          + #math.equation(block: false, alt: "the fraction 1 over 10")[$display(frac(1, 10))$]
+ #math.equation(block: false, alt: "the fraction π over 10")[$display(frac(π, 10))$] sq km
+ #math.equation(block: false, alt: "the fraction π over 5")[$display(frac(π, 5))$] km


        
      
      
        #figure(figph[circle], alt: "circle", caption: none)


      
      
        #figure(figph[circle], alt: "circle", caption: none)


        
          + #math.equation(block: false, alt: "the fraction 5 over 6")[$display(frac(5, 6))$]
+ #math.equation(block: false, alt: "the fraction 15 π over 2")[$display(frac(15 π, 2))$] sq m
+ #math.equation(block: false, alt: "5 π")[$5 π$] m


        
      
      
        #figure(figph[circle], alt: "circle", caption: none)


      
      
        South America stretches across #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] of longitude at the equator, from Quito in Ecuador to the east coast of Brazil.  (The figure at right shows a view of the earth from above the north pole.)  The radius of the earth is about 3960 miles.  How wide is South America at the equator?


        #figure(figph[equator], alt: "equator", caption: none)


        
          2070 miles


        
      
      
        The radius of the earth is about 3960 miles.  What distance will you cover if you travel north by one degree of latitude?  (See the figure at right.)  There are #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] of latitude from the south pole to the north pole.


        #figure(figph[earth], alt: "earth", caption: none)


      
      
        The moon is about 240,000 miles from earth.  It moves in a roughly circular orbit, completing one revolution in 28 days.


        + How far does the moon move around the earth in one day?
+ What is the speed of the moon relative to the earth, in miles per hour?


        #figure(figph[earth and moon], alt: "earth and moon", caption: none)


        
          + 54,000 miles
+ 2240 mph


        
      
      
        The earth is about 93,000,000 miles from the sun, and its orbit is approximately circular.


        + How far does the earth move around the sun in one month?
+ What is the speed of the earth relative to the sun, in miles per day?  (Assume a month has 30 days.)


        #figure(figph[earth and sun], alt: "earth and sun", caption: none)


      
      
        + Use the distance formula to write an equation for the circle of radius 6 centered at the point #math.equation(block: false, alt: "open parenthesis 3 , minus 2 close parenthesis")[$( 3 , − 2 )$].
+ Use the distance formula to derive an equation for the circle of radius #math.equation(block: false, alt: "r")[$r$] centered at the point #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$].


        
          + #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared plus open parenthesis y plus 2 close parenthesis squared equals 36")[$( x − 3 )^(2) + ( y + 2 )^(2) = 36$]
+ #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")[$( x − h )^(2) + ( y − k )^(2) = r^(2)$]


        
      
      
        Find the coordinates of the indicated points on each circle.


        + #math.equation(block: false, alt: "open parenthesis x minus 4 close parenthesis squared plus open parenthesis y minus 5 close parenthesis squared equals 9")[$( x − 4 )^(2) + ( y − 5 )^(2) = 9$]
            #figure(figph[circle], alt: "circle", caption: none)
+ #math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared equals 25")[$( x + 3 )^(2) + ( y − 1 )^(2) = 25$]#figure(figph[circle], alt: "circle", caption: none)