Answer
$$r = 4\cos \theta $$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} - 4x = 0 \cr
& {\text{Convert to polar form, using }}x = r\cos \theta ,{\text{ }}y = r\sin \theta \cr
& {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2} - 4\left( {r\cos \theta } \right) = 0 \cr
& {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta - 4r\cos \theta = 0 \cr
& {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) - 4r\cos \theta = 0 \cr
& {r^2} - 4r\cos \theta = 0 \cr
& r - 4\cos \theta = 0 \cr
& r = 4\cos \theta \cr
& {\text{The equation represents a circle}} \cr
& {\text{Graph}} \cr} $$
