# How do you find the center and radius of the circle given 3x^2+3y^2+12x-6y+9=0?

Nov 23, 2016

The center and radius are related to the coefficient of the equation.

#### Explanation:

First divide all the equation by three to put in simple form:

${x}^{2} + {y}^{2} + 4 x - 2 y + 3 = 0$

Now consider the general equation of the circle is:

${\left(x - {x}_{o}\right)}^{2} + {\left(y - {y}_{o}\right)}^{2} - {R}^{2} = 0$

where the center is the point $\left({x}_{o} , {y}_{o}\right)$ and $R$ is the radius.

Expanding:

${x}^{2} - 2 x {x}_{o} + {x}_{o}^{2} + {y}^{2} - 2 y {y}_{o} + {y}_{o}^{2} - {R}^{2} = 0$

Comparing similar terms, you can easily see that:

$- 2 {x}_{o} = 4 \implies {x}_{o} = - 2$
$- 2 {y}_{o} = - 2 \implies {y}_{o} = 1$
${x}_{o}^{2} + {y}_{o}^{2} - {R}^{2} = 3 \implies R = \sqrt{2}$

graph{x^2+y^2+4x-2y+3=0 [-10.375, 9.625, -4.08, 5.92]}