# How do you find the center and radius for x^2+y^2+10x+6y-27=0?

Jun 22, 2016

The center is $\left(- 5 , - 3\right)$ and radius is $\sqrt{61}$.

#### Explanation:

If the equation is in the form

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

The center is $\left(h , k\right)$ and radius is $r$.

Hence, let us convert the equation ${x}^{2} + {y}^{2} + 10 x + 6 y - 27 = 0$ to this form,

${x}^{2} + 10 x + {y}^{2} + 6 y = 27$

$\Leftrightarrow {x}^{2} + 10 x + 25 + {y}^{2} + 6 y + 9 = 27 + 25 + 9$

$\Leftrightarrow {\left(x + 5\right)}^{2} + {\left(y + 3\right)}^{2} = 61$

$\Leftrightarrow {\left(x - \left(- 5\right)\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {\left(\sqrt{61}\right)}^{2}$

Hence, the center is $\left(- 5 , - 3\right)$ and radius is $\sqrt{61}$.