# How do you find the center and radius for x^2+ y^2 +10x-8y-41=0?

Jun 29, 2016

center (-5;4); radius $2 \sqrt{82}$

#### Explanation:

The center is (-a/2;-b/2)

and the radius is $\frac{1}{2} \sqrt{{a}^{2} + {b}^{2} - 4 c}$

when the equation of a circle is

${x}^{2} + {y}^{2} + a x + b y + c = 0$

so, in this case,

a=10; b=-8; c=-41

therefore the center is (-5;4)

and the radius is $\frac{1}{2} \sqrt{100 + 64 + 164} = \sqrt{328} = 2 \sqrt{82}$