# How do you find the center and radius of the circle x^2-12x+y^2+4y+15=0?

Dec 18, 2016

The center is $\left(6 , - 2\right)$ and the radius $= 5$

#### Explanation:

We complete the squares and rearrange the equation

${x}^{2} - 12 x + {y}^{2} + 4 y = - 15$

${x}^{2} - 12 x + 36 + {y}^{2} + 4 y + 4 = - 15 + 36 + 4$

And now we factorise

${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 25 = {5}^{2}$

We compare this equation to the standard equation of a circle

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

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

In our case,

The center is $\left(6 , - 2\right)$ and the radius $= 5$

graph{x^2-12x+y^2+4y+15=0 [-6.22, 16.275, -6.95, 4.3]}