Answer
$ (x)^2+(y-4)^2=16\ (y\ne0)$
a circle with center $(0,4)$, radius $r=4$, 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 $r\ csc\theta=8\ (\theta\ne0)\rightarrow r=8sin\theta\rightarrow r^2=8r\ sin\theta \rightarrow x^2+y^2=8y \rightarrow (x)^2+(y-4)^2=16\ (y\ne0)$
2. We can identify the above equation as a circle with center $(0,4)$, radius $r=4$, with a hole $(0,0)$.
3. See graph.
