Answer
$(x)^2+(y+1)^2=1\ (y\ne0)$
a circle with center $(0,-1)$, radius $r=1$, hole $(0,0)$.
See graph.
Work Step by Step
1. Use the formula $r^2=x^2+y^2, r\ sin\theta=y$, we have $r\ csc\theta=-2\ (\theta\ne0)\rightarrow r=-2sin\theta\rightarrow r^2=-2r\ sin\theta \rightarrow x^2+y^2=-2y \rightarrow (x)^2+(y+1)^2=1\ (y\ne0)$
2. We can identify the above equation as a circle with center $(0,-1)$, radius $r=1$, and a hole $(0,0)$.
3. See graph.
