# How do you find the center and radius of the circle 6x^2 + 6y^2 + 12x − y = 0?

Jan 12, 2017

C(-1;1/12)

and $r = \frac{\sqrt{145}}{12}$

#### Explanation:

You would write the given equation in the standard form:

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

Then it is, by dividing all terms by 6,

${x}^{2} + {y}^{2} + 2 x - \frac{1}{6} y = 0$

You would find the center and radius by using the following formulas:

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

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

Then C(-2/2;-(-1/6)/2)=(-1;1/12)

and $r = \frac{1}{2} \sqrt{{2}^{2} + {\left(- \frac{1}{6}\right)}^{2} - 4 \cdot 0} = \frac{1}{2} \sqrt{4 + \frac{1}{36}} = \frac{1}{2} \sqrt{\frac{145}{36}} = \frac{\sqrt{145}}{12}$