Answer
$ (x-2)^2+(y)^2=4\ (x\ne0)$
a circle with center $(2,0)$, radius $r=2$, hole $(0,0)$.
See graph.
Work Step by Step
1. Use the formula $r^2=x^2+y^2, r\ cos\theta=x$, we have $r\ sec\theta=4\ (\theta\ne\frac{\pi}{2})\rightarrow r=4cos\theta\rightarrow r^2=4r\ cos\theta \rightarrow x^2+y^2=4x \rightarrow (x-2)^2+(y)^2=4\ (x\ne0)$
2. We can identify the above equation as a circle with center $(2,0)$, radius $r=2$, with a hole $(0,0)$.
3. See graph.
