# How do you find the center and radius of the circle given x^2+y^2+9x-8y+4=0?

Dec 18, 2016

The center is $\left(- \frac{9}{2} , 4\right)$ and the radius $r = \frac{\sqrt{129}}{2}$

#### Explanation:

We complete the squares and rearrange the equation

${x}^{2} + 9 x + {y}^{2} - 8 y = - 4$

${x}^{2} + 9 x + \frac{81}{4} + {y}^{2} - 8 y + 16 = - 4 + \frac{81}{4} + 16$

And now we factorise

${\left(x + \frac{9}{2}\right)}^{2} + {\left(y - 4\right)}^{2} = \frac{129}{4}$

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 $\left(- \frac{9}{2} , 4\right)$ and the radius $= \frac{\sqrt{129}}{2}$