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

Jul 24, 2016

Do a double completion of square, with respect to $x$ and $y$, in order to convert to the arm ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = r$.

$1 \left({x}^{2} + 2 x\right) + 1 \left({y}^{2} + 6 y\right) = 6$

$1 \left({x}^{2} + 2 x + 1 - 1\right) + 1 \left({y}^{2} + 6 y + 9 - 9\right) = 6$

$1 \left({x}^{2} + 2 x + 1\right) - 1 + 1 \left({y}^{2} + 6 y + 9\right) - 9 = 6$

${\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} - 10 = 6$

${\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} = 16$

In the form ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = r$, the radius is given by $\sqrt{r}$ and the centre is at $\left(a , b\right)$. Then, the centre is at $\left(- 1 , - 3\right)$ and the radius measures $4$ units.

Hopefully this helps!