# How do you find the center and radius for x^2 + y^2 - 8x - 2y - 8 = 0?

Jun 1, 2016

The center is $\left(4 , 1\right)$ and the radius $3$.

#### Explanation:

$0 = {x}^{2} + {y}^{2} - 8 x - 2 y - 8$

$= {x}^{2} - 8 x + 16 + {y}^{2} - 2 y + 1 - 9$

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

Add ${3}^{2}$ to both ends and transpose to get:

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

This is in the form:

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

which is the general form of the equation of a circle with centre $\left(h , k\right) = \left(4 , 1\right)$ and radius $r = 3$