# How do you find the center and radius for y^2+ x^2-12y +10x+ 57=0?

Jul 2, 2016

#### Explanation:

complete the square -- TWICE!!

${y}^{2} + {x}^{2} - 12 y + 10 x + 57 = 0$

${y}^{2} - 12 y + {x}^{2} + 10 x + 57 = 0$

${\left(y - 6\right)}^{2} - 36 + {\left(x + 5\right)}^{2} - 25 + 57 = 0$

${\left(y - 6\right)}^{2} + {\left(x + 5\right)}^{2} = 36 + 25 - 57$

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