# How do you find the center and radius of the circle x^2 + y^2 -4x+12y-36 = 0?

Sep 30, 2016

The center is $\left(2 , - 6\right)$ and the radius is $2 \sqrt{19}$

#### Explanation:

${x}^{2} + {y}^{2} - 4 x + 12 y - 36 = 0$

Rearrange the terms by grouping the x terms, grouping the y terms, and moving the constant term to the other side
${x}^{2} - 4 x \textcolor{w h i t e}{a a a} + {y}^{2} + 12 y \textcolor{w h i t e}{a a a} = 36 \textcolor{w h i t e}{a a a}$

Complete the square for x terms and the y terms.

To complete the square for the x terms, divide the coefficient of the "middle term" $\textcolor{red}{4} x$ by 2, square the result to get $\textcolor{b l u e}{4}$ and add $\textcolor{b l u e}{4}$ to both sides of the equation.
${x}^{2} - \textcolor{red}{4} x + \textcolor{b l u e}{4} \textcolor{w h i t e}{a a a} + {y}^{2} + 12 y \textcolor{w h i t e}{a a} = 36 + \textcolor{b l u e}{4}$

To complete the square for the y terms, divide the coefficient of the "middle term" $\textcolor{red}{12} y$ by 2, square the result to get $\textcolor{b l u e}{36}$ and add $\textcolor{b l u e}{36}$ to both sides of the equation.
${x}^{2} - 4 x + 4 \textcolor{w h i t e}{a a a}$$+ {y}^{2} + \textcolor{red}{12} y + \textcolor{b l u e}{36} = 36 + 4 + \textcolor{b l u e}{36}$

${x}^{2} - 4 x + 4 \textcolor{w h i t e}{a a a} + {y}^{2} + 12 y + 36 = 76$

Factor each group of terms.
$\left(x - 2\right) \left(x - 2\right) + \left(y + 6\right) \left(y + 6\right) = 76$

${\left(x - 2\right)}^{2} + {\left(y + 6\right)}^{2} = 76$

This equation of a circle is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
where $\left(h , k\right)$ is the center and $r$ is the radius.

Thus the center is $\left(2 , - 6\right)$ and the radius is $\sqrt{76}$ or $2 \sqrt{19}$