Answer
$(x-3)^{2}+y^{2}=9$
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=6\cos\theta$
...Substitute $\cos\theta$ with $\displaystyle \frac{x}{r}$,
$r=6\displaystyle \cdot\frac{x}{r}\qquad/\times r$
$r^{2}=6x$
...Substitute $x^{2}+y^{2}$ for $r^{2}$,
$x^{2}+y^{2}=6x$
$(x^{2}-6x+9-9)+y^{2}=0$
... completing the square$, +9-9=0$ ...
$(x^{2}-6x+9)+y^{2}=9$
$(x-3)^{2}+y^{2}=9$
