Answer
$r=3\cos \theta$
Work Step by Step
Consider $x^2+y^2-3x=0$
or, $(x-\dfrac{3}{2})^2+y^2=(\dfrac{9}{4})$
Here, we get the equation of a circle with center $(\dfrac{3}{2},0)$ and radius $\dfrac{3}{2}$
This implies that
$r^2-3r \cos \theta=0 \implies r^2=3r \cos \theta$
Hence, $r=3\cos \theta$
