Answer
$x^{2}+y^{2}=y+\sqrt{x^{2}+y^{2}} \quad \text{or} \quad x^2+y^2-y-\sqrt{x^2+y^2}=0$
Work Step by Step
Multiply $r$ to both sides of the equation to obtain
\begin{align*}
r\cdot r &=r(\sin\theta+1)
\\r^{2}&=r\sin\theta+r
\end{align*}
Since $r^{2}=x^{2}+y^{2}$ and $r\sin\theta=y$, then
\begin{align*}
r^2&=r\sin\theta+r\\
x^2+y^2&=y+\sqrt{x^2+y^2}\\
x^2+y^2-\left(y+\sqrt{x^2+y^2}\right)&=0\\
x^2+y^2-y-\sqrt{x^2+y^2}&=0\\
\end{align*}
