# How do you find the center and radius of the circle 9x^2+9y^2+6y-107=0?

##### 1 Answer
Jun 17, 2016

C(0;-1/3); $r = 2 \sqrt{3}$

#### Explanation:

you can write the equivalent equation by dividing all terms by 9:

${x}^{2} + {y}^{2} + \frac{2}{3} y - \frac{107}{9} = 0$

you obtain the center of the circle by using the formulas:

C(-a/2;-b/2)

and radius by

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

where a, b, c are in the following general circle equation:

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

So
C(0;-2/3*1/2)

C(0;-1/3)

$r = \frac{1}{2} \sqrt{\frac{4}{9} - 4 \left(- \frac{107}{9}\right)}$

$r = \frac{1}{2} \sqrt{\frac{4}{9} + \frac{428}{9}}$

$r = \frac{1}{2} \sqrt{\frac{432}{9}}$

$r = \frac{1}{2} \sqrt{48}$

$r = \frac{1}{2} \cdot 4 \sqrt{3}$

$r = 2 \sqrt{3}$