What is the polar form of the equation?
(x+2)^2+y^2=4
r^2=4rcosθ
r(r+4sinθ)=0
r(r+4cosθ)=0
r^2=4rsinθ
(x + 2)^2 + y^2 = 4 x = r cos θ y = rsin θ
( rcosθ + 2 )^2 + (rsin θ)^2 = 4
r^2cos^2θ + 4rcosθ+ 4 + r^2sin^2θ = 4
r^2 ( sin^2θ + cos^2θ) + 4r cosθ = 0 sin^2θ + cos^2θ = 1
r^2 + 4rcosθ = 0
r ( r + 4cosθ) = 0
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