# How do you find the center and radius of the following circle x^2 + 6x +y^2 -2y +6=0?

Jun 23, 2016

Complete the square twice to find that the center is $\left(- 3 , 1\right)$ and the radius is $2$.

#### Explanation:

The standard equation for a circle is:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
Where $\left(h , k\right)$ is the center and $r$ is the radius.

We want to get ${x}^{2} + 6 x + {y}^{2} - 2 y + 6 = 0$ into that format so we can identity the center and radius. To do so, we need to complete the square on the $x$ and $y$ terms separately. Starting with $x$:
$\left({x}^{2} + 6 x\right) + {y}^{2} - 2 y + 6 = 0$
$\left({x}^{2} + 6 x + 9\right) + {y}^{2} - 2 y + 6 = 9$
${\left(x + 3\right)}^{2} + {y}^{2} - 2 y + 6 = 9$

Now we can go ahead and subtract $6$ from both sides:
${\left(x + 3\right)}^{2} + {y}^{2} - 2 y = 3$

We are left to complete the square on the $y$ terms:
${\left(x + 3\right)}^{2} + \left({y}^{2} - 2 y\right) = 3$
${\left(x + 3\right)}^{2} + \left({y}^{2} - 2 y + 1\right) = 3 + 1$
${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = 4$

The equation of this circle is therefore ${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = 4$. Note this can be rewritten as ${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - \left(1\right)\right)}^{2} = 4$, so the center $\left(h , k\right)$ is $\left(- 3 , 1\right)$. The radius is found by taking the square root of the number on the right side of the equation (which, in this case, is $4$). Doing so yields a radius of $2$.