# How do you find the center and radius of the circle x^2 + y^2 - 10x - 8y - 8 = 0?

Jun 29, 2016

The centre is $\left(5 , 4\right)$, radius $r = 7.$

#### Explanation:

The given eqn. :${x}^{2} + {y}^{2} - 10 x - 8 y - 8 = 0.$
Completing squares on $L . H . S .$, we get,
${x}^{2} - 10 x + 25 + {y}^{2} - 8 y + 16 = 8 + 25 + 16 = 49.$
$\therefore {\left(x - 5\right)}^{2} + {\left(y - 4\right)}^{2} = {7}^{2.} \ldots \ldots \ldots \ldots \ldots \left(1\right)$

Now the std. eqn. of a circle with centre $\left(h , k\right)$ & radius$= r$ is given by,

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

Comparing $\left(1\right)$ with the std. eqn, we find that,
the centre is $\left(5 , 4\right)$, radius $r = 7.$