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

Dec 29, 2015

There are few ways which can be used to find the center and radius, one of the way is using completion of squares, and is given below in a step by step manner.

#### Explanation:

The given equation
${x}^{2} + {y}^{2} + 10 x + 8 y = 8$

One way to do the problem is to use completing the square to get the problem in the form

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

Where $\left(h , k\right)$ is the center and $r$ is the radius.

${x}^{2} + {y}^{2} + 10 x + 8 y = 8$

Let us group the $x$ and $y$ together.

${x}^{2} + 10 x + {y}^{2} + 8 y = 8$

Divide $x$ coefficient by $2$ square and add on both sides, do the same for $y$ coefficient.

${x}^{2} + 10 x + {\left(5\right)}^{2} + {y}^{2} + 8 y + {\left(4\right)}^{2} = 8 + {5}^{2} + {4}^{2}$
${\left(x + 5\right)}^{2} + {\left(y + 4\right)}^{2} = 8 + 25 + 16$
${\left(x + 5\right)}^{2} + {\left(y + 4\right)}^{2} = 49$
${\left(x - \left(- 5\right)\right)}^{2} + {\left(y - \left(- 4\right)\right)}^{2} = {7}^{2}$

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

We can see the center $\left(h , k\right)$ is $\left(- 5 , - 4\right)$
Radius $r$ is $7$