Answer
$x^2+(y-1)^2 = 1$
We can see the graph below:
Work Step by Step
$r = 2~sin~\theta$
We know that $r = \sqrt{x^2+y^2}$
Since $y = r~sin~\theta$, then $sin~\theta = \frac{y}{\sqrt{x^2+y^2}}$
We can find the rectangular coordinates:
$r = 2~sin~\theta$
$\sqrt{x^2+y^2} = \frac{2y}{\sqrt{x^2+y^2}}$
$x^2+y^2 = 2y$
$x^2+y^2 - 2y = 0$
$x^2+(y-1)^2 = 1$
We can see the graph below:
