What is the polar form of the equation?
(x+2)^2+y^2=4
r(r+4sinθ)=0
r(r+4cosθ)=0
r2=4rcosθ
r2=4rsinθ
+23246 x = r·cos(theta) y = r·sin(theta)
(x + 2)2 + y2 = 4 ---> ( r·cos(theta) + 2 )2 + ( r·sin(theta) )2 = 4
---> r2·cos2(theta) + 4·r·cos(theta) + 4 + r2·sin2(theta) = 4 multiply out
---> r2·cos2(theta) + 4·r·cos(theta) + r2·sin2(theta) = 0 subtract the 4
---> r2·cos2(theta) + r2·sin2(theta) + 4·r·cos(theta) = 0 rearrange
---> r2·( cos2(theta) + sin2(theta) ) + 4·r·cos(theta) = 0 factor
---> r2 + 4·r·cos(theta) = 0 sin2 + cos2 = 1
---> r( r + 4·cos(theta) ) = 0 factor
