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Study Guides > College Algebra

Circles

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.
Definition: Circle A circle with center (h,k) and radius r > 0 is the set of all points (x,y) in the plane whose distance to (h,k) is r.
Image of a circle with center (h,k) and radius , r, with a labeled point on the edge of the circle, (x,y) From the picture, 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:

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

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

The Standard Equation of a Circle

The equation of a circle with center (h,k) and raduis r > 0 is

(x−h)2+(y−k)2=r2{\left(x - h\right)}^{2}+{\left(y - k\right)}^{2}={r}^{2}\\

Example 1: Write the Standard Equation of a Circle given the center and a radius.

Write the standard equation of the circle whose center is ( -2,3) and whose radius is 5.

Solution

Here, (h,k) = (-2, 3) and r = 5, so we get:

(x−(−2))2+(y−3)2=(5)2(x+2)2+(y−3)2=25\begin{array}{rrr}{\left(x-\left(-2\right)\right)}^{2}+{\left(y-3\right)}^{2}& \hfill = &\hfill{\left(5\right)}^{2}\\{\left(x+2\right)}^{2}+{\left(y-3\right)}^{2}& \hfill = &\hfill{25}\end{array}\\

Example 2: Graph a circle given an equation in standard form.

Graph (x+2)2+(y−1)2=4{\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}={ 4 }\\. Find the center and radius.

Solution

From the standard equation of a circle, we have that (x+2){\left(x+2\right)}\\ is (x−h){\left(x-h\right)}\\, so h = -2 and (y−1){\left(y-1\right)}\\ is (y−k){\left(y-k\right)}\\ so k = 1. This tells us that our center is (-2, 1). Furthermore, r2=4{r}^{2} = {4}\\, so r = 2. Graphing gives us
Graph of a circle with radius 2 centered at (-2, 1) on the cartesian coordinate axes. Graph of a circle with radius 2 centered at (-2, 1) on the cartesian coordinate axes.
If we were to expand the equation in the previous example and gather up like terms, instead of the easily recognizable (x+2)2+(y−1)2=4{\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}={ 4 }\\, we'd be contending with x2+4x+y2−2y+1=0{x}^{2} +4x+{y}^{2}-2y+1 = 0\\.  If we're given such an equation, we can complete the square in each of the variables to see if it fits the form given by the standard form for the equation of a circle by following the steps below.
Write the Equation of a Circle in Standard Form
  1. Group the same variables together on one side of the equation and position the constant on the other side.
  2. Complete the square on both variables as needed.
  3. Divide both sides by the coefficient of the squares. (For circles, they will be the same).

Example 3: Complete the square to find the center and radius of a circle whose equation is not in standard form.

Find the center and radius of the circle whose equation is 3x2−6x+3y2+4y−4=03{x}^{2}-6x+3{y}^{2} +4y-4=0\\

Solution

3x2−6x+3y2+4y−4=0 3x2−6x+3y2+4y=4 add 4 to both sides 3(x2−2x+1)+3(y2+43y)=4 factor out leading coefficients 3(x2−2x+1)+3(y2+43y+49)=4+3(1)+3(49) complete the square in x, y 3(x−1)2+3(y+23)2=253 factor (x−1)2+(y+23)2=259 divide both sides by 3 \begin{array}{cccc}3{x}^{2}-6x+3{y}^{2} + 4y - 4 &\hfill = & \hfill 0 & \text{ }\\3{x}^{2}-6x+3{y}^{2} + 4y &\hfill=& \hfill 4 & \text{ add 4 to both sides }\\3\left({x}^{2} - 2x + 1\right) + 3\left({y}^{2} +\frac{4}{3}y\right) &\hfill = & \hfill 4 & \text{ factor out leading coefficients }\\3\left({x}^{2} - 2x + 1\right) + 3\left({y}^{2} +\frac{4}{3}y + \frac{4}{9}\right) &\hfill = & \hfill 4 +3(1) +3\left(\frac{4}{9}\right) & \text{ complete the square in x, y }\\3{\left(x -1\right)}^{2} + 3{\left(y +\frac{2}{3}\right)}^{2} &\hfill = & \hfill \frac{25}{3} & \text{ factor }\\{\left(x -1\right)}^{2} + {\left(y +\frac{2}{3}\right)}^{2} &\hfill = & \hfill \frac{25}{9} & \text{ divide both sides by 3 }\end{array}\\ From the standard equation of a circle, we can identify x-1 as x-h, so h = 1,and y+23y + \frac{2}{3}\\ as y-k, so k=−23k = -\frac{2}{3}\\. Hence, the center is (h,k) = (1,−23)\left(1,-\frac{2}{3}\right)\\. Furthermore we see that r2=259{r}^{2} = \frac{25}{9}\\ so the radius is 53\frac{5}{3}\\.
It is possible to obtain equations like (x−3)2+(y+1)2=0{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}={ 0 }\\ or (x−3)2+(y+1)2=−1{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}={ -1 }\\, neither of which describes a circle. Can you think of why this is true? Consider what points, if any, would lie on a line described by one of the two equations.  What value defines the radius of each circle, and does it make sense?   In our next example, the Midpoint Formula is used in conjunction with the ideas presented above to write the standard equation of a circle given the endpoints that define it's diameter. To review the Midpoint Formula, go to this page.  

Example 4: Given the endpoints of it's diameter, find the equation of a circle using the Midpoint Formula.

Write the standard equation of a circle given that it's diameter has the endpoints (-1,3) and (2,4).

Solution

We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields
Circle graphed on the Cartesian Coordinate axes passing through the points (-1,3)and (2,4). Circle graphed on the Cartesian Coordinate axes passing through the points (-1,3)and (2,4).
Since the circle passes through the endpoints of the diameter, we know that the midpoint (h , k) of the diameter is the center of the circle. From the midpoint formula, we get

M=(h,k)=(x1+x22,y1+y22) =(−1+22,3+42) =(12,72)\begin{array}{ccc}M= (h, k) &\hfill= &\hfill\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\\\text{ }&\hfill=&\hfill\left(\frac{-1+2}{2},\frac{3+4}{2}\right)\\\text{ }&\hfill=&\hfill\left(\frac{1}{2},\frac{7}{2}\right)\end{array}\\

The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus,

r=12(x2−x1)2+(y2−y1)2 =12(2−(−1))2+(4−3)2 =1232+12 =102\begin{array}{lll}{ r }&\hfill=&\hfill\frac{1}{2}\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2}+{\left({y}_{2} - {y}_{1}\right)}^{2}}\\\text{ }&\hfill=&\hfill\frac{1}{2}\sqrt{{\left(2 - (-1)\right)}^{2}+{\left(4 - 3\right)}^{2}}\\\text{ }&\hfill=&\hfill\frac{1}{2}\sqrt{{3}^{2}+{1}^{2}}\\\text{ }&\hfill=&\hfill\frac{\sqrt{10}}{2}\end{array}\\

Finally, since (102)2=104{\left(\frac{\sqrt{10}}{2}\right)}^{2} = \frac{10}{4}\\, our answer becomes (x−12)2+(y−72)2=104{\left(x-\frac{1}{2}\right)}^{2}+{\left(y-\frac{7}{2}\right)}^{2}=\frac{10}{4}\\
We close this section with the most important circle in all of mathematics: the Unit Circle.  
Definition: Unit Circle A circle with center at (0,0) and radius of 1. The standard equation of the Unit Circle is x2+y2=1{x}^{2} + {y}^{2} = 1\\
 

Example 5: Identify points on the Unit Circle.

Find the points on the Unit Circle whose y-coordinate is 32\frac{\sqrt{3}}{2}\\.

Solution

Replace y with 32\frac{\sqrt{3}}{2}\\in the equation x2+y2=1{x}^{2} + {y}^{2} = 1\\ to get

x2+y2=1x2+(32)2=1x2+34=1x2=14x=±14x=±12\begin{array}{ccc}\hfill{x}^{2}+{y}^{2}&\hfill=&\hfill{ 1 }\\\hfill{x}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2} &\hfill= &\hfill{ 1 }\\\hfill{x}^{2}+\frac{3}{4}&\hfill=&\hfill{ 1 }\\\hfill{x}^{2} &\hfill= &\hfill{\frac{1}{4}}\\\hfill{x} &\hfill = &\pm\sqrt{\frac{1}{4}}\\\hfill{x} &=&\pm{\frac{1}{2}}\end{array}\\

The resulting points on the Unit Circle whose y-coordinate is 32\frac{\sqrt{3}}{2}\\ are (12,32) and (−12,32)\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\text{ and }\left(\frac{-1}{2},\frac{\sqrt{3}}{2}\right)\\

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