Answer
$ r=-2\cos\theta$
Work Step by Step
The circle passes through the origin, as (0,0) satisfies the Cartesian equation.
Completing the square,
$(x^{2}+2x+1)+y^{2}=0+1$
$(x+1)^{2}+y^{2}=1$
The radius is $1$, and the center is at $(-1,0)$.
In polar coordinates, the center lies at $r_{0}=1, \theta_{0}=\pi,\qquad P(1,\pi)$
A circle passing through the origin, of radius $a$, centered at $P_{0}(r_{0}, \theta_{0}),$
has the polar equation
$r=2a\cos(\theta-\theta_{0})$
So this circle has equation
$r=2(1)\cos(\theta-\pi)$
We apply the identity
$\cos(\theta-\pi)=-\cos\theta$,
in which case the equation is
$ r=-2\cos\theta$
