Answer
$ r=-\sin\theta$
Work Step by Step
The circle passes through the origin, as (0,0) satisfies the Cartesian equation.
Completing the square gives:
$x^{2}+(y^{2}+y+(\displaystyle \frac{1}{2})^{2})+y^{2}=0+(\frac{1}{2})^{2}$
$x^{2}+(y+\displaystyle \frac{1}{2})^{2}=\frac{1}{4}$
The radius is $\displaystyle \frac{1}{2}$, and the center is at $(0,-\displaystyle \frac{1}{2})$.
In polar coordinates, the center lies at $r_{0}=\displaystyle \frac{1}{2}, \theta_{0}=\pi/2,\qquad P(\frac{1}{2},\pi/2)$
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 the equation
$r=2(\displaystyle \frac{1}{2})\cos(\theta-\frac{\pi}{2})$
or
$r=\displaystyle \cos(\theta-\frac{\pi}{2})$
Using the identity
$\displaystyle \cos(\theta\pm\frac{\pi}{2})=\mp\sin\theta$,
We get:
$ r=-\sin\theta$
