# How do you find the center and radius for x^2 + y^2 + 3x + 6y +9=0?

Aug 5, 2016

$C \left(- \frac{3}{2} , - 3\right)$; $r = \frac{3}{2}$

#### Explanation:

The standard form of the equation of the circle is:

${x}^{2} + {y}^{2} + a x + b y + c = 0$

The center is obtained by:

$C \left(- \frac{a}{2} , - \frac{b}{2}\right)$

$r = \frac{1}{2} \sqrt{{a}^{2} + {b}^{2} - 4 c}$

Then let's apply the to the case:

${x}^{2} + {y}^{2} + 3 x + 6 y + 9 = 0$

and have

$C \left(- \frac{3}{2} , - 3\right)$

$r = \frac{1}{2} \sqrt{9 + \cancel{36} - \cancel{36}} = \frac{3}{2}$