8.3 Circles

Our next entry in the conic sections menagerie is the circle. Recall from Geometry that a circle can be determined by fixing a point (called the center) and a positive number (called the radius) as follows.

Coordinate-plane figure. — **Figure 8.46**

From the diagram, we see that a point $(x, y)$ is on the circle if and only if its distance to $(h, k)$ is $r$ . We express this relationship algebraically using the Distance Formula, Equation A.1, as

$r = \sqrt{{(x - h)}^{2} + {(y - k)}^{2}}$

By squaring both sides of this equation, we get an equivalent equation (since $r greater than 0$ ) which gives us the standard equation of a circle.

Equation

The Standard Equation of a Circle:

The equation of a circle with center $(h, k)$ and radius $r greater than 0$ is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2} .$

Note in the standard equation of a circle, both of the variables squared. This is a quick way to distinguish the equation of a circle from that of a parabola in which only one of the variables is squared.

We put Equation 8.3 to good use in the following example.

Example 1

For each of the equations below:
- Graph the equation in the $x y$ -plane.
- Find the center and radius.
1. ${(x + 2)}^{2} + {(y - 1)}^{2} = 4$
2. $3 x^{2} - 6 x + 3 y^{2} + 4 y - 4 = 0$
Graph $f (x) = - \sqrt{4 x - x^{2}}$ .
Find the standard form of the circle satisfying the following characteristics:
1. The points $open parenthesis minus 1, 3 close parenthesis$ and $open parenthesis 2, 4 close parenthesis$ are the endpoints of a diameter.
2. The circle whose graph is below.
  Figure 8.47

1. Rewriting ${(x + 2)}^{2} + {(y - 1)}^{2} = 4$ as ${(x - (- 2))}^{2} + {(y - 1)}^{2} = {(2)}^{2}$ , we identify $h = - 2$ , $k equals 1$ and $r equals 2$ . Thus we have a circle centered at $open parenthesis minus 2, 1 close parenthesis$ with a radius of $2$ .
  To help us create a detailed graph, we start from the center $open parenthesis minus 2, 1 close parenthesis$ and move two units to the left and two units up and down to the right to identify four points on the graph.¹ We get $(- 2 - 2, 1) = (- 4, 1)$ , $(- 2 + 2, 1) = (0, 1)$ , $(- 2, 1 + 2) = (- 2, 3)$ and $(- 2, 1 - 2) = (- 2, - 1)$ . Our graph is below on the left.
2. In order to make use of Equation 8.3, we need to put $3 x^{2} - 6 x + 3 y^{2} + 4 y - 4 = 0$ into standard form. To that end, we complete the square on both the $x$ and $y$ terms and collect the constants to the other side of the equation as demonstrated below.
  
  From Equation 8.3, we identify $h equals 1$ , $k = - \frac{2}{3}$ , and $r = \frac{5}{3}$ . Hence, we have a circle with center $(1, - \frac{2}{3})$ and radius $\frac{5}{3}$ .
  As above, we find four points on the circle by starting at the center $(1, - \frac{2}{3})$ and moving up, down, to the left, and to the right $\frac{5}{3}$ units. Doing so produces the following points: $(- \frac{2}{3}, - \frac{2}{3})$ , $(\frac{8}{3}, - \frac{2}{3})$ , $open parenthesis 1, 1 close parenthesis$ , and $(1, - \frac{7}{3})$ . Our graph is below on the right.
  Figure 8.48 The graph of ${(x + 2)}^{2} + {(y - 1)}^{2} = 4$
  Figure 8.49 The graph of $3 x^{2} - 6 x + 3 y^{2} + 4 y - 4 = 0$
We are asked to graph $f (x) = - \sqrt{4 x - x^{2}}$ , which, at first glance, seems out of place in this section. However, to graph means we graph the equation $y = - \sqrt{4 x - x^{2}}$ .
Squaring both sides, we get $y^{2} = {(- \sqrt{4 x - x^{2}})}^{2}$ or $y^{2} = 4 x - x^{2}$ . Rearranging this equation gives $x^{2} - 4 x + y^{2} = 0$ . Completing the square, we obtain ${(x - 2)}^{2} + y^{2} = 4$ which, when rewritten as ${(x - 2)}^{2} + {(y - 0)}^{2} = {(2)}^{2}$ is precisely the standard form of a circle as written in Equation 8.3.
With $h equals 2$ , $k equals 0$ and $r equals 2$ , we know the graph of ${(x - 2)}^{2} + y^{2} = 4$ is a circle of radius $2$ centered at $open parenthesis 2, 0 close parenthesis$ . However, the graph we want isn't the entire circle.² Indeed, we want the graph of $y = - \sqrt{4 x - x^{2}}$ . Because of the ` ', we want the lower semicircle, graphed below on the left.
1. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. We plot the data given to us below on the right.
  Figure 8.50 The graph of $f (x) = - \sqrt{4 x - x^{2}}$ .
  Figure 8.51
  Since the given points are endpoints of a diameter, we know their midpoint $(h, k)$ is the center of the circle. Likewise, the diameter of the circle is the distance between the given points, so we can find the radius of the circle by taking half of this distance. Using Equations A.2 and A.1, respectively, we get:
  $\begin{matrix} (h, k) & = & (\frac{x_{0} + x_{1}}{2}, \frac{y_{0} + y_{1}}{2}) \\ = & (\frac{- 1 + 2}{2}, \frac{3 + 4}{2}) \\ = & (\frac{1}{2}, \frac{7}{2}) \end{matrix}$
  $\begin{matrix} r & = & \frac{1}{2} \sqrt{{(x_{1} - x_{0})}^{2} + {(y_{1} - y_{0})}^{2}} \\ = & \frac{1}{2} \sqrt{{(2 - (- 1))}^{2} + {(4 - 3)}^{2}} \\ = & \frac{1}{2} \sqrt{3^{2} + 1^{2}} \\ = & \frac{\sqrt{10}}{2} \end{matrix}$
  Finally, since ${(\frac{\sqrt{10}}{2})}^{2} = \frac{10}{4} = \frac{5}{2}$ , our answer becomes ${(x - \frac{1}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{5}{2}$
2. From the graph given to us, we are safe to assume the center of the circle is $open parenthesis minus 2, 2 close parenthesis$ since the circle appears to be tangent to the coordinate axes at $open parenthesis minus 2, 0 close parenthesis$ and $open parenthesis 0, 2 close parenthesis$ .³ Moreover, since the distance from $open parenthesis minus 2, 2 close parenthesis$ to either of $open parenthesis minus 2, 0 close parenthesis$ or $open parenthesis 0, 2 close parenthesis$ is $2$ , the radius of the circle is $2$ . Per Equation 8.3, our answer is ${(x - (- 2))}^{2} + {(y - 2)}^{2} = {(2)}^{2}$ or ${(x + 2)}^{2} + {(y - 2)}^{2} = 4$ .

In number above, we needed to transform a given equation into the standard form as stated in Equation 8.3. We record these steps below. Note that given an equation that represents a circle, both variables need to be squared and the squared terms must have the same coefficients.

To Write the Equation of a Circle in Standard Form

Group common variables together on one side of the equation and put the constant on the other.
Complete the square on both variables as needed.
Divide both sides by the coefficient of the squares. (For circles, they will be the same.)

It is possible to obtain equations like ${(x - 3)}^{2} + {(y + 1)}^{2} = 0$ or ${(x - 3)}^{2} + {(y + 1)}^{2} = - 1$ , neither of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if any, points lie on the graphs of these two equations.

We close this section with a brief discussion of the so-called Unit Circle.⁴

In some ways, we may think of the Unit Circle as the progenitor of all circles. Indeed, if we divide both sides of Equation 8.3 by $r^{2}$ , we obtain the alternate standard form of a circle below.

Equation

The Alternate Standard Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r greater than 0$ is

$\frac{{(x - h)}^{2}}{r^{2}} + \frac{{(y - k)}^{2}}{r^{2}} = 1$

Taking this one step further, we may rewrite Equation 8.4 as

${(\frac{x - h}{r})}^{2} + {(\frac{y - k}{r})}^{2} = 1 .$

Hence, every circle can be obtained from the Unit Circle via the transformations discussed in Section 5.4.⁵

Our last example has us find some important points on the the Unit Circle.

Example 2

Find the points on the unit circle with $y$ -coordinate $\frac{\sqrt{3}}{2}$ .

Note that all points $(x, y)$ on the Unit Circle satisfy the equation $x^{2} + y^{2} = 1$ . Hence, our first step is to replace $y$ with $\frac{\sqrt{3}}{2}$ and solve for $x$ .

$\begin{matrix} x^{2} + y^{2} & = & 1 \\ x^{2} + {(\frac{\sqrt{3}}{2})}^{2} & = & 1 \\ \frac{3}{4} + x^{2} & = & 1 \\ x^{2} & = & \frac{1}{4} \\ x & = & \pm \sqrt{\frac{1}{4}} & extract square roots \\ x & = & \pm \frac{1}{2} \end{matrix}$

We find $x = \pm \frac{1}{2}$ so our final answers are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ and $(- \frac{1}{2}, \frac{\sqrt{3}}{2})$ .

Exercises

In Exercises -, graph the circle in the $x y$ -plane. Find the center and radius.

${(x + 1)}^{2} + {(y + 5)}^{2} = 100$
${(x - 4)}^{2} + {(y + 2)}^{2} = 9$
${(x + 3)}^{2} + {(y - \frac{7}{13})}^{2} = \frac{1}{4}$
${(x - 5)}^{2} + {(y + 9)}^{2} = {(\ln (8))}^{2}$
${(x + e)}^{2} + {(y - \sqrt{2})}^{2} = π^{2}$
${(x - π)}^{2} + {(y - e^{2})}^{2} = 91^{\frac{2}{3}}$

In Exercises -, complete the square in order to put the equation into standard form. Identify the center and the radius or explain why the equation does not represent a circle.⁶

$x^{2} - 4 x + y^{2} + 10 y = - 25$
$- 2 x^{2} - 36 x - 2 y^{2} - 112 = 0$
$3 x^{2} + 3 y^{2} + 24 x - 30 y - 3 = 0$
$x^{2} + y^{2} + 5 x - y - 1 = 0$
$x^{2} + x + y^{2} - \frac{6}{5} y = 1$
$4 x^{2} + 4 y^{2} - 24 y + 36 = 0$
For each of the odd numbered equations given in Exercises -, find two or more explicit functions of $x$ represented by each of the equations. (See Example 8.2.2 in Section 8.2.)

In Exercises -, graph each function by recognizing it as a semicircle.

$f (x) = \sqrt{4 - x^{2}}$
$g (x) = - \sqrt{6 x - x^{2}}$
$f (x) = - \sqrt{3 - 2 x - x^{2}}$
$g (x) = - 2 + \sqrt{9 - x^{2}}$

In Exercises -, find an equation for the circle or semicircle whose graph is given.

Figure 8.52
Figure 8.53
Figure 8.54 $x$ , $y$ -intercept $open parenthesis 0, 0 close parenthesis$
Figure 8.55 $x$ , $y$ -intercept $open parenthesis 0, 0 close parenthesis$

In Exercises -, find the standard equation of the circle which satisfies the given criteria.

center $open parenthesis 3, 5 close parenthesis$ , passes through $(- 1, - 2)$
center $open parenthesis 3, 6 close parenthesis$ , passes through $open parenthesis minus 1, 4 close parenthesis$
endpoints of a diameter: $open parenthesis 3, 6 close parenthesis$ and $open parenthesis minus 1, 4 close parenthesis$
endpoints of a diameter: $(\frac{1}{2}, 4)$ , $(\frac{3}{2}, - 1)$
The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet.⁷ Find an equation for the wheel assuming that its center lies on the $y$ -axis and that the ground is the $x$ -axis.
Verify that the following points lie on the Unit Circle:
$(\pm 1, 0)$ , $(0, \pm 1)$ , $(\pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2})$ , $(\pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2})$ and $(\pm \frac{\sqrt{3}}{2}, \pm \frac{1}{2})$
Discuss with your classmates how to obtain the alternate standard equation of a circle, Equation 8.4, from the equation of the Unit Circle, $x^{2} + y^{2} = 1$ using the transformations discussed in Section 5.4. (Thus every circle is just a few transformations away from the Unit Circle.)
Find a one-to-one function whose graph is half of a circle.
HINT: Think piecewise …

Answers

Center $(- 1, - 5)$ , radius $10$
Figure 8.56
Center $open parenthesis 4, minus 2 close parenthesis$ , radius $3$
Figure 8.57
Center $(- 3, \frac{7}{13})$ , radius $\frac{1}{2}$
Figure 8.58
Center $open parenthesis 5, minus 9 close parenthesis$ , radius $ln ⁡ open parenthesis 8 close parenthesis$
Figure 8.59
Center $(- e, \sqrt{2})$ , radius $π$
Figure 8.60
Center $(π, e^{2})$ , radius $the cube root of 91$
Figure 8.61
${(x - 2)}^{2} + {(y + 5)}^{2} = 4$ Center $open parenthesis 2, minus 5 close parenthesis$ , radius $r equals 2$
${(x + 9)}^{2} + y^{2} = 25$ Center $open parenthesis minus 9, 0 close parenthesis$ , radius $r equals 5$
${(x + 4)}^{2} + {(y - 5)}^{2} = 42$ Center $open parenthesis minus 4, 5 close parenthesis$ , radius $r = \sqrt{42}$
${(x + \frac{5}{2})}^{2} + {(y - \frac{1}{2})}^{2} = \frac{30}{4}$ Center $(- \frac{5}{2}, \frac{1}{2})$ , radius $r = \frac{\sqrt{30}}{2}$
${(x + \frac{1}{2})}^{2} + {(y - \frac{3}{5})}^{2} = \frac{161}{100}$ Center $(- \frac{1}{2}, \frac{3}{5})$ , radius $r = \frac{\sqrt{161}}{10}$
$x^{2} + {(y - 3)}^{2} = 0$ This is not a circle.
For number:
- $f (x) = - 5 + \sqrt{99 - 2 x - x^{2}}$ represents the upper semicircle.
- $g (x) = - 5 - \sqrt{99 - 2 x - x^{2}}$ represents the lower semicircle.
For number:
- $f (x) = \frac{7}{13} + \frac{1}{2} \sqrt{- 4 x^{2} - 24 x - 35}$ represents the upper semicircle.
- $g (x) = \frac{7}{13} - \frac{1}{2} \sqrt{- 4 x^{2} - 24 x - 35}$ represents the lower semicircle.
For number:
- $f (x) = \sqrt{2} + \sqrt{π^{2} - e^{2} - 2 e x - x^{2}}$ represents the upper semicircle.
- $g (x) = \sqrt{2} - \sqrt{π^{2} - e^{2} - 2 e x - x^{2}}$ represents the lower semicircle.
For number:
- $f (x) = - 5 + \sqrt{4 x - x^{2}}$ represents the upper semicircle.
- $g (x) = - 5 - \sqrt{4 x - x^{2}}$ represents the lower semicircle.
For number:
- $f (x) = 5 + \sqrt{26 - 8 x - x^{2}}$ represents the upper semicircle.
- $g (x) = 5 - \sqrt{26 - 8 x - x^{2}}$ represents the lower semicircle.
For number:
- $f (x) = \frac{3}{5} + \frac{1}{5} \sqrt{34 - 25 x - 25 x^{2}}$ represents the upper semicircle.
- $g (x) = \frac{3}{5} - \frac{1}{5} \sqrt{34 - 25 x - 25 x^{2}}$ represents the lower semicircle.
$f (x) = \sqrt{4 - x^{2}}$
Figure 8.62
$g (x) = - \sqrt{6 x - x^{2}}$
Figure 8.63
$f (x) = - \sqrt{3 - 2 x - x^{2}}$
Figure 8.64
$g (x) = - 2 + \sqrt{9 - x^{2}}$
Figure 8.65
${(x - 1)}^{2} + y^{2} = 9$
${(x - 4)}^{2} + {(y - 4)}^{2} = 16$
$y = 4 - \sqrt{16 - x^{2}}$
$y = \sqrt{8 x - x^{2}}$
${(x - 3)}^{2} + {(y - 5)}^{2} = 65$
${(x - 3)}^{2} + {(y - 6)}^{2} = 20$
${(x - 1)}^{2} + {(y - 5)}^{2} = 5$
${(x - 1)}^{2} + {(y - \frac{3}{2})}^{2} = \frac{13}{2}$
$x^{2} + {(y - 72)}^{2} = 4096$

Note the center of the circle is not on the graph of the circle! ↩
For one thing, the graph of a circle fails the Vertical Line Test so it does not represent $y$ as a function of $x$ in this case. ↩
Recall that for every point $P$ on the circle, the tangent line at $P$ is perpendicular to the radial line containing the center and $P$ . Since the circle is tangent to the $x$ -axis at $open parenthesis minus 2, 0 close parenthesis$ , the center must lie on a line perpendicular to the $x$ -axis which contains $open parenthesis minus 2, 0 close parenthesis$ or $x = - 2$ . Likewise, the circle is tangent to the $y$ -axis at $open parenthesis 0, 2 close parenthesis$ , the center must lie on $y equals 2$ . Hence the center is $open parenthesis minus 2, 2 close parenthesis$ . ↩
Widely regarded as the most important circle in all of mathematics. ↩
See Exercise . ↩
…assuming the equation were graphed in the $x y$ -plane. ↩
Source: Cedar Point's webpage . ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.