# How would you find the center and radius of x^2 + y^2 + 12x - 6y = 0?

Jun 1, 2018

An circle with radius $3 \sqrt{5}$, and centre $\left(- 6 , 3\right)$

#### Explanation:

We have:

${x}^{2} + {y}^{2} + 12 x - 6 y = 0$

First we can collect "like" terms:

$\left\{{x}^{2} + 12 x\right\} + \left\{{y}^{2} - 6\right\} = 0$

Next we complete the square independently on the $x$ and $y$ terms

$\therefore \left\{{\left(x + 6\right)}^{2} - {6}^{2}\right\} + \left\{{\left(y - 3\right)}^{2} - {3}^{2}\right\} = 0$

$\therefore {\left(x + 6\right)}^{2} - 36 + {\left(y - 3\right)}^{2} - 9 = 0$

$\therefore {\left(x + 6\right)}^{2} + {\left(y - 3\right)}^{2} = 36 + 9$

$\therefore {\left(x + 6\right)}^{2} + {\left(y - 3\right)}^{2} = 45$

$\therefore {\left(x + 6\right)}^{2} + {\left(y - 3\right)}^{2} = {\left(3 \sqrt{5}\right)}^{2}$

And as such, we can identify the conic as an circle with radius $3 \sqrt{5}$, and centre $\left(- 6 , 3\right)$

graph{x^2 + y^2 + 12x - 6y = 0 [-24.5, 15.5, -6.32, 13.68]}