# How do you find the center and radius of the circle x^2+y^2+8x+15=0?

Sep 1, 2016

Center is $\left(- 4 , 0\right)$ and radius is $1$.

#### Explanation:

${x}^{2} + {y}^{2} + 8 x + 15 = 0$ can be written as

${x}^{2} + 8 x + {y}^{2} = - 15$ or

${x}^{2} + 8 x + 16 + {y}^{2} = - 15 + 16$ or

${\left(x + 4\right)}^{2} + {\left(y - 0\right)}^{2} = {1}^{2}$ or

${\left(x - \left(- 4\right)\right)}^{2} + {\left(y - 0\right)}^{2} = {1}^{2}$

which is the locus of a point which moves so that it is always at a distance of $1$ from point $\left(- 4 , 0\right)$.

Hence, this represents a circle whose center is $\left(- 4 , 0\right)$ and radius is $1$.