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