# How do you find the center and radius of a circle using a polynomial (x^2) + (y^2) + 6x - 4y = 12?

Feb 8, 2016

Re-write the equation in standard form for a circle to get:
center: $\left(- 3 , 2\right)$ and radius: $5$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} + 6 x - 4 y = 12$

Complete the square for each of the $x$ and $y$ components:
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} + 6 x \textcolor{red}{+ 9}\right) + \left({y}^{2} - 4 y \textcolor{b l u e}{+ 4}\right) = 12 \textcolor{red}{+ 9} \textcolor{b l u e}{+ 4}$

Re-write as squared binomials and squared constant:
$\textcolor{w h i t e}{\text{XXX}} {\left(x \textcolor{g r e e n}{+ 3}\right)}^{2} + {\left(y - \textcolor{\mathmr{and} a n \ge}{2}\right)}^{2} = {\textcolor{c y a n}{5}}^{2}$

Since the standard form for a circle with center $\left(\textcolor{g r e e n}{a} , \textcolor{\mathmr{and} a n \ge}{b}\right)$ and radius $\textcolor{c y a n}{r}$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{g r e e n}{a}\right)}^{2} + {\left(y - \textcolor{\mathmr{and} a n \ge}{b}\right)}^{2} = {\textcolor{c y a n}{r}}^{2}$

The given circle has a center $\left(\textcolor{g r e e n}{- 3} , \textcolor{\mathmr{and} a n \ge}{2}\right)$ and radius $\textcolor{c y a n}{5}$
graph{x^2+y^2+6x-4y=12 [-11.955, 8.045, -3, 7]}