Asked by Luke
what would x^2+y^2-4x+2y=0 (Equation for a cirlce) be in
r^2=(x-h)^2(y-k)^2 form?
tnx
I will start you off...
x^2+y^2-4x+2y=0
x^2-4x+y^2 +2y =0
x^2-4x+ 4 + y^2 +2y+ 4 =+ 4+ 4
(x^2-4x+ 4) + (y^2 +2y+ 4) =8
r^2=(x-h)^2(y-k)^2 form?
tnx
I will start you off...
x^2+y^2-4x+2y=0
x^2-4x+y^2 +2y =0
x^2-4x+ 4 + y^2 +2y+ 4 =+ 4+ 4
(x^2-4x+ 4) + (y^2 +2y+ 4) =8
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