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

Question

3256 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-29T11:49:32

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

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

and the radius is $1/2sqrt(a^2+b^2-4c)$

when the equation of a circle is

$x^2+y^2+ax+by+c=0$

so, in this case,

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

therefore the center is $(-5;4)$

and the radius is $1/2sqrt(100+64+164)=sqrt(328)=2sqrt(82)$

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