Skip Navigation

Circles in the Coordinate Plane

Standard equation based on center and radius.

Atoms Practice
This indicates how strong in your memory this concept is
Practice Now
Turn In
Circles in the Coordinate Plane

Circles in the Coordinate Plane

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 \begin{align*}(x, y)\end{align*} is a point on the circle, then the distance from the center to this point would be the radius, \begin{align*}r\end{align*}. \begin{align*}x\end{align*} is the horizontal distance and \begin{align*}y\end{align*} is the vertical distance. This forms a right triangle. From the Pythagorean Theorem, the equation of a circle centered at the origin is \begin{align*}x^2+y^2=r^2\end{align*}.

The center does not always have to be on (0, 0). If it is not, then we label the center \begin{align*}(h, k)\end{align*}. 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 \begin{align*}(h, k)\end{align*} and radius \begin{align*}r\end{align*} is \begin{align*}r^2=(x-h)^2+(y-k)^2\end{align*}.

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?


Example 1

Find the center and radius of the following circle.


Rewrite the equation as \begin{align*}(x-(-2))^2+(y-5)^2=7^2\end{align*}. The center is (-2, 5) and \begin{align*}r = 7\end{align*}.

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.

Example 2

Find the center and radius of the following circle.

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

First plug in the center to the standard equation.

\begin{align*}(x-4)^2+(y-(-1))^2&=r^2 \\ (x-4)^2+(y+1)^2&=r^2\end{align*}

Now, plug in (-1, 2) for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and solve for \begin{align*}r\end{align*}.

\begin{align*}(-1-4)^2+(2+1)^2&=r^2\\ (-5)^2+(3)^2&=r^2\\ 25+9&=r^2\\ 34&=r^2\end{align*}

Substituting in 34 for \begin{align*}r^2\end{align*}, the equation is \begin{align*}(x-4)^2+(y+1)^2=34\end{align*}.

Example 3

Graph \begin{align*}x^2+y^2=9\end{align*}.

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 4

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 \begin{align*}r = 6\end{align*}. Plugging this into the equation of a circle, we get: \begin{align*}(x-(-3))^2+(y-3)^2=6^2\end{align*} or \begin{align*}(x+3)^2+(y-3)^2=36\end{align*}.

Example 5

Determine if the following points are on \begin{align*}(x+1)^2+(y-5)^2=50\end{align*}.

Plug in the points for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in \begin{align*}(x+1)^2+(y-5)^2=50\end{align*}.

  1. (8, -3)
    \begin{align*}(8+1)^2+(-3-5)^2 &= 50\\ 9^2+(-8)^2 &= 50\\ 81+64 & \ne 50\end{align*}

 (8, -3) is not on the circle

  1. (-2, -2)
    \begin{align*}(-2+1)^2+(-2-5)^2 &= 50\\ (-1)^2+(-7)^2 &= 50\!\\ 1+49 &= 50\end{align*}

 (-2, -2) is on the circle


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

  1. \begin{align*}(x+5)^2+(y-3)^2=16\end{align*}
  2. \begin{align*}x^2+(y+8)^2=4\end{align*}
  3. \begin{align*}(x-7)^2+(y-10)^2=20\end{align*}
  4. \begin{align*}(x+2)^2+y^2=8\end{align*}

Find the equation of the circles below.

  1. Is (-7, 3) on \begin{align*}(x+1)^2+(y-6)^2=45\end{align*}?
  2. Is (9, -1) on \begin{align*}(x-2)^2+(y-2)^2=60\end{align*}?
  3. Is (-4, -3) on \begin{align*}(x+3)^2+(y-3)^2=37\end{align*}?
  4. Is (5, -3) on \begin{align*}(x+1)^2+(y-6)^2=45\end{align*}?

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)

Review (Answers)

To see the Review answers, open this PDF file and look for section 9.12. 

Notes/Highlights Having trouble? Report an issue.

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


circle The set of all points that are the same distance away from a specific point, called the center.
radius The distance from the center to the outer rim of a circle.
Distance Formula The distance between two points (x_1, y_1) and (x_2, y_2) can be defined as d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.
Origin The origin is the point of intersection of the x and y axes on the Cartesian plane. The coordinates of the origin are (0, 0).

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Circles in the Coordinate Plane.
Please wait...
Please wait...