Answer
(a) See the graph.
(b) $(x^2+y^2-3x)^2=9(x^2+y^2)$
Work Step by Step
(b)
$x=r~cos~θ$
$y=r~sin~θ$
$r^2=x^2+y^2$
$r=3+3~cos~θ$
$r-3=3~\frac{x}{r}$
$r^2-3r=3x$
$r^2-3x=3r~~$ (Square both sides)
$(r^2-3x)^2=9r^2$
$(x^2+y^2-3x)^2=9(x^2+y^2)$
