How do you find the center and the radius of the circle; the graph the circle of x^2+y^2+8x-8y-93=0?

Jan 20, 2016

The centre is $\left(- 4 , 4\right)$ and the radius is $5 \sqrt{5}$

Explanation:

The standard form of a circle is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

To get this equation into the right form, start by grouping like terms.
${x}^{2} + 8 x + {y}^{2} - 8 y = 93$
${\left(x + 4\right)}^{2} - 16 + {\left(y - 4\right)}^{2} - 16 = 93$
${\left(x + 4\right)}^{2} + {\left(y - 4\right)}^{2} = 93 + 32 = 125$

${\left(x + 4\right)}^{2} + {\left(y - 4\right)}^{2} = {\left(5 \sqrt{5}\right)}^{2}$

The centre is $\left(- 4 , 4\right)$ and the radius is $5 \sqrt{5}$