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

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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)#