Chapter 10 - Practice Exercises - Page 594: 35

$r=3\cos \theta$

Here, we have $x^2+y^2-3x=0$ This implies that $(x-\dfrac{3}{2})^2+y^2=\dfrac{9}{4}$ Thus, we have an equation of a circle with center $(\dfrac{3}{2},0)$ and radius $\dfrac{3}{2}$ So, $r^2-3r \cos \theta=0$ Hence, $r=3\cos \theta$