# How do you find the center and radius of the circle given x^2+y^2+8x-6y=0?

Jan 12, 2017

This is the equation of a circle, center $\left(- 4 , 3\right)$ and radius $= 5$

#### Explanation:

We need

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

Complete the squares for the x and y

${x}^{2} + {y}^{2} + 8 x - 6 y = 0$

${x}^{2} + 8 x + {y}^{2} - 6 y = 0$

${x}^{2} + 8 x + \textcolor{red}{{\left(\frac{8}{2}\right)}^{2}} + {y}^{2} - 6 y + \textcolor{b l u e}{{\left(\frac{6}{2}\right)}^{2}} = \textcolor{red}{{\left(\frac{8}{2}\right)}^{2}} + \textcolor{b l u e}{{\left(\frac{6}{2}\right)}^{2}}$

${x}^{2} + 8 x + \textcolor{red}{{\left(4\right)}^{2}} + {y}^{2} - 6 y + \textcolor{b l u e}{{\left(3\right)}^{2}} = \textcolor{red}{{\left(4\right)}^{2}} + \textcolor{b l u e}{{\left(3\right)}^{2}}$

${\left(x + 4\right)}^{2} + {\left(y - 3\right)}^{2} = 25 = {5}^{2}$

This is the equation of a circle, center $\left(- 4 , 3\right)$ and radius $= 5$

graph{(x^2+y^2+8x-6y)=0 [-15.25, 13.23, -3.42, 10.82]}