Answer
$9(x^{2}+y^{2})=(x^{2}+y^{2}+3x)^{2}$
Work Step by Step
Formulas relating rectangular $(x,y)$ and polar $(r,\theta)$ coordinates:
$ x=r\cos\theta$ and $ y=r\sin\theta$
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$ r=3-3\cos\theta$
... multiply both sides with r,
$ r^{2}=3r-3r\cos\theta$
... substitute $r^{2}$ and $ r\cos\theta$ using the formulas
$x^{2}+y^{2}=3r-3x\qquad /+3x$
... isolate r on one side,
$ 3r=x^{2}+y^{2}+3x\qquad$... square both sides,
$9r^{2}=(x^{2}+y^{2}+3x)^{2}\qquad $... substitute $r^{2}$
$9(x^{2}+y^{2})=(x^{2}+y^{2}+3x)^{2}$,
... an equation in rectangular coordinates.