# What Is the Equation of a Circle?

Rule

### TheCircle

The equation of a circle with its center at $\left({x}_{0},{y}_{0}\right)$ and with radius $r$ is given by

 ${\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}={r}^{2}$

This means that all values for $\left(x,y\right)$ that fulfill the equation are on the circle boundary.

To find the radius of the circle you can use this formula:

 $r=\sqrt{{\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}}$

Example 1

You are working with a circle with its center at $\left(-1,2\right)$. The point $\left(3,5\right)$ is on the circle boundary. What is the radius of this circle?

Here you have $\left({x}_{0},{y}_{0}\right)=\left(-1,2\right)$ and the point on the circle is $\left(x,y\right)=\left(3,5\right)$. The radius of the circle is therefore

 $r=\sqrt{{\left(3+1\right)}^{2}+{\left(5-2\right)}^{2}}=5$

Example 2

Show that ${x}^{2}+6x+{y}^{2}-2y-6=0$ is a circle

To show this, you have to use the method to complete the square. You add and subtract $\phantom{\rule{-0.17em}{0ex}}{\left(\frac{b}{2}\right)}^{2}$ for both the $x$-terms and the $y$-terms. You then get the following:

$\begin{array}{llll}\hfill 0& ={x}^{2}+6x+{y}^{2}-2y-6,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+6x+\phantom{\rule{-0.17em}{0ex}}{\left(\frac{6}{2}\right)}^{2}-\phantom{\rule{-0.17em}{0ex}}{\left(\frac{6}{2}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+{y}^{2}-2y+\phantom{\rule{-0.17em}{0ex}}{\left(\frac{-2}{2}\right)}^{2}-\phantom{\rule{-0.17em}{0ex}}{\left(\frac{-2}{2}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+6x+9-9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+{y}^{2}-2y+1-1-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x+3\right)}^{2}+{\left(y-1\right)}^{2}-9-1-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x+3\right)}^{2}+{\left(y-1\right)}^{2}-16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 0& ={x}^{2}+6x+{y}^{2}-2y-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+6x+\phantom{\rule{-0.17em}{0ex}}{\left(\frac{6}{2}\right)}^{2}-\phantom{\rule{-0.17em}{0ex}}{\left(\frac{6}{2}\right)}^{2}+{y}^{2}-2y+\phantom{\rule{-0.17em}{0ex}}{\left(\frac{-2}{2}\right)}^{2}-\phantom{\rule{-0.17em}{0ex}}{\left(\frac{-2}{2}\right)}^{2}-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+6x+9-9+{y}^{2}-2y+1-1-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x+3\right)}^{2}+{\left(y-1\right)}^{2}-9-1-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x+3\right)}^{2}+{\left(y-1\right)}^{2}-16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Thus,
 ${\left(x+3\right)}^{2}+{\left(y-1\right)}^{2}={4}^{2}$

You can see from this that ${x}^{2}+6x+{y}^{2}-2y-6=0$ is a circle with its center at $\left(-3,1\right)$ and a radius $r=4$.