# How do you find the center and radius of the circle given x^2+y^2+4x-8=0?

Oct 22, 2016

we have the centre as: $\left(- 2 , 0\right)$

radius $\sqrt{12} = 2 \sqrt{3}$

#### Explanation:

The standard equation of a circle is

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

where $\left(a , b\right)$ is the centre, and $r$ is the radius,

we thus have to complete the square on the given equation.

${x}^{2} + {y}^{2} + 4 x - 8 = 0$

rearrange slightly

${x}^{2} + 4 x + {y}^{2} - 8 = 0$

$\left({x}^{2} + 4 x + {2}^{2}\right) + {y}^{2} - 8 - {2}^{2} = 0$

${\left(x + 2\right)}^{2} + {y}^{2} - 12 = 0$

${\left(x + 2\right)}^{2} + {y}^{2} = 12$

comparing with

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

we have the centre as: $\left(- 2 , 0\right)$

radius $\sqrt{12} = 2 \sqrt{3}$

graph{x^2+y^2+4x-8=0 [-10, 10, -5, 5]}