Answer
$(x^{2}+y^{2}-2y)^{2}=x^{2}+y^{2}$
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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$r=1+2\sin\theta|$
... multiply both sides with r,
$ r^{2}=r+2r\sin\theta$
... substitute $r^{2}$ and $ r\sin\theta$ using the formulas
$x^{2}+y^{2}=r+2y\qquad /-2y$
$ x^{2}+y^{2}-2y=r\qquad$... square both sides,
$(x^{2}+y^{2}-2y)^{2}=r^{2}\qquad $... substitute $r^{2}$
$(x^{2}+y^{2}-2y)^{2}=x^{2}+y^{2}$,
... an equation in rectangular coordinates.