r2cosθ−arcos2θ−2a2cosθ=0r2cosθ−ar(2cos2θ−1)−2a2cosθ=0r2cosθ−2arcos2θ+ar−2a2cosθ=0rcosθ(r−2acosθ)+a(r−2acosθ)=0(rcosθ+a)(r−2acosθ)=0⇒rcosθ+a=0,r−2acosθ=0
Converting rcosθ+a=0 to Cartesian coordinates
⇒x+a=0
which represents a straight line
Also r−2acosθ=0
Multiplying by r on both sides
⇒r2−2arcosθ=0⇒(√x2+y2)2−2ax=0⇒x2+y2−2ax=0
which represents a circle .
Hence proved that the given equation represents a circle and a straight line