# How do you find the center and radius of the circle -2x^2 - 2y^2 + 24x - 54 = 0?

Dec 3, 2016

This is a circle, center $\left(6 , 0\right)$ and radius $3$

#### Explanation:

Rearrange the equation to look like

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

This is a circle center $\left(a , b\right)$ and radius $r$

Dividing by $- 2$

${x}^{2} - 12 x + {y}^{2} = - 27$

Completing the squares

${x}^{2} - 12 x + 36 + {y}^{2} = - 27 + 36$

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

so this is a circle, center $\left(6 , 0\right)$ and radius $3$

graph{-2x^2-2y^2+24x-54=0 [-2.64, 13.16, -3.416, 4.49]}