<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

9.7: Extension: Writing and Graphing the Equations of Circles

Difficulty Level: At Grade Created by: CK-12
Turn In

Learning Objectives

  • Graph a circle.
  • Find the equation of a circle in the xy plane.
  • Find the radius and center, given the equation of a circle and vice versa.
  • Find the equation of a circle, given the center and a point on the circle.

Graphing a Circle in the Coordinate Plane

Recall that the definition of a circle is the set of all points that are the same distance from the center. This definition can be used to find an equation of a circle in the coordinate plane.

Let’s start with the circle centered at (0, 0). If (x,y) is a point on the circle, then the distance from the center to this point would be the radius, r. x is the horizontal distance y is the vertical distance. This forms a right triangle. From the Pythagorean Theorem, the equation of a circle, centered at the origin is x2+y2=r2.

Example 1: Graph x2+y2=9.

Solution: The center is (0, 0). It’s radius is the square root of 9, or 3. Plot the center, and then go out 3 units in every direction and connect them to form a circle.

The center does not always have to be on (0, 0). If it is not, then we label the center (h,k) and would use the distance formula to find the length of the radius.


If you square both sides of this equation, then we would have the standard equation of a circle.

Standard Equation of a Circle: The standard equation of a circle with center (h,k) and radius r is r2=(xh)2+(yk)2.

Example 2: Find the center and radius of the following circles.

a) (x3)2+(y1)2=25

b) (x+2)2+(y5)2=49


a) Rewrite the equation as (x3)2+(y1)2=52. The center is (3, 1) and r=5.

b) Rewrite the equation as (x(2))2+(y5)2=72. The center is (-2, 5) and r=7.

When finding the center of a circle always take the opposite sign of what the value is in the equation.

Example 3: Find the equation of the circle below.

Solution: First locate the center. Draw in the horizontal and vertical diameters to see where they intersect.

From this, we see that the center is (-3, 3). If we count the units from the center to the circle on either of these diameters, we find r=6. Plugging this into the equation of a circle, we get: (x(3))2+(y3)2=62 or (x+3)2+(y3)2=36.

Finding the Equation of a Circle

Example 4: Determine if the following points are on (x+1)2+(y5)2=50.

a) (8, -3)

b) (-2, -2)

Solution: Plug in the points for x and y in (x+1)2+(y5)2=50.

a) (8+1)2+(35)2=5092+(8)2=5081+6450

(8, -3) is not on the circle

b) (2+1)2+(25)2=50(1)2+(7)2=501+49=50

(-2, -2) is on the circle

Example 5: Find the equation of the circle with center (4, -1) and passes through (-1, 2).

Solution: First plug in the center to the standard equation.


Now, plug in (-1, 2) for x and y and solve for r.


Substituting in 34 for r2, the equation is (x4)2+(y+1)2=34.

Review Questions

  • Questions 1-4 are similar to Examples 1 and 2.
  • Questions 5-8 are similar to Example 3.
  • Questions 9-11 are similar to Example 4.
  • Questions 12-15 are similar to Example 5.

Find the center and radius of each circle. Then, graph each circle.

  1. (x+5)2+(y3)2=16
  2. x2+(y+8)2=4
  3. (x7)2+(y10)2=20
  4. (x+2)2+y2=8

Find the equation of the circles below.

  1. Is (-7, 3) on (x+1)2+(y6)2=45?
  2. Is (9, -1) on (x2)2+(y2)2=60?
  3. Is (-4, -3) on (x+3)2+(y3)2=37?
  4. Is (5, -3) on (x+1)2+(y6)2=45?

Find the equation of the circle with the given center and point on the circle.

  1. center: (2, 3), point: (-4, -1)
  2. center: (10, 0), point: (5, 2)
  3. center: (-3, 8), point: (7, -2)
  4. center: (6, -6), point: (-9, 4)

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Show Hide Details
Files can only be attached to the latest version of section
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original