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Circles in the Coordinate Plane

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What if you were given the length of the radius of a circle and the coordinates of its center? How could you write the equation of the circle in the coordinate plane? After completing this Concept, you'll be able to write the standard equation of a circle.

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Graphing Circles CK-12

James Sousa: Write the Standard Form of a Circle


Recall that a circle is the set of all points in a plane 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 and y is the vertical distance. This forms a right triangle. From the Pythagorean Theorem, the equation of a circle centered at the origin is x^2+y^2=r^2 .

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


If you square both sides of this equation, then you would have the standard equation of a circle. The standard equation of a circle with center (h, k) and radius r is r^2=(x-h)^2+(y-k)^2 .

Example A

Graph x^2+y^2=9 .

The center is (0, 0). Its radius is the square root of 9, or 3. Plot the center, plot the points that are 3 units to the right, left, up, and down from the center and then connect these four points to form a circle.

Example B

Find the equation of the circle below.

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+(y-3)^2=6^2 or (x+3)^2+(y-3)^2=36 .

Example C

Determine if the following points are on (x+1)^2+(y-5)^2=50 .

a) (8, -3)

b) (-2, -2)

Plug in the points for x and y in (x+1)^2+(y-5)^2=50 .

a) (8+1)^2+(-3-5)^2 &= 50\\9^2+(-8)^2 &= 50\\81+64 & \ne 50

(8, -3) is not on the circle

b) (-2+1)^2+(-2-5)^2 &= 50\\(-1)^2+(-7)^2 &= 50\!\\1+49 &= 50

(-2, -2) is on the circle

Graphing Circles CK-12

Guided Practice

Find the center and radius of the following circles.

  1. (x-3)^2+(y-1)^2=25
  2. (x+2)^2+(y-5)^2=49

3. Find the equation of the circle with center (4, -1) and which passes through (-1, 2).


1. Rewrite the equation as (x-3)^2+(y-1)^2=5^2 . The center is (3, 1) and r = 5 .

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

Keep in mind that, due to the minus signs in the formula, the coordinates of the center have the opposite signs of what they may initially appear to be.

3. First plug in the center to the standard equation.

(x-4)^2+(y-(-1))^2&=r^2 \\(x-4)^2+(y+1)^2&=r^2

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


Substituting in 34 for r^2 , the equation is (x-4)^2+(y+1)^2=34 .


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

  1. (x+5)^2+(y-3)^2=16
  2. x^2+(y+8)^2=4
  3. (x-7)^2+(y-10)^2=20
  4. (x+2)^2+y^2=8

Find the equation of the circles below.

  1. Is (-7, 3) on (x+1)^2+(y-6)^2=45 ?
  2. Is (9, -1) on (x-2)^2+(y-2)^2=60 ?
  3. Is (-4, -3) on (x+3)^2+(y-3)^2=37 ?
  4. Is (5, -3) on (x+1)^2+(y-6)^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)

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