# How do you find the center and radius for x^2+y^2+x-6y+9=0?

Oct 28, 2016

center = $\left(- \frac{1}{2} , 3\right)$

radius =$\frac{1}{2}$

#### Explanation:

We have the equation,

${x}^{2} + {y}^{2} + x - 6 y + 9 = 0$

When a circle is in this form we must complete the square in order to put in a usable format. like,

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

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

So rearrange,

${x}^{2} + x + {y}^{2} - 6 y = - 9$

And complete the square,

${\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4} + {\left(y - 3\right)}^{2} - 9 = - 9$

Rearrange,

${\left(x + \frac{1}{2}\right)}^{2} + {\left(y - 3\right)}^{2} = \frac{1}{4}$

So from this equation, we know that our circle centre is at,

$\left(- \frac{1}{2} , 3\right)$

and the radius of our circle is,

$\sqrt{\frac{1}{4}} = \frac{1}{2}$