How do you find the center and radius of the circle with the given equation #x^2 + y^2 +4x - 8y - 44 = 0#?

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Answer 1 · 2016-08-02T15:51:05

The Centre is #(-2,4)# and radius#=8#.

Let us rewrite the given eqn. of circle as,

# x^2+4x+y^2-8y=44#, and, complete the squares on L.H.S..

#:. x^2+4x+4+y^2-8y+16=44+4+16=64#.

#:. (x+2)^2+(y-4)^2=8^2#.

We compare this eqn. with # : (x-h)^2+(y-k)^2=r^2#, representing the

circle with centre #(h,k)# and radius #r#.

Therefore, the Centre is #(-2,4)# and radius#=8#.