Answer
$ r=\displaystyle \frac{4}{3}\sin\theta$
Work Step by Step
The circle passes through the origin, as (0,0) satisfies the Cartesian equation.
Completing the square,
$x^{2}+(y^{2}-2(\displaystyle \frac{2}{3})y+(\frac{2}{3})^{2})+y^{2}=0+(\frac{2}{3})^{2}$
$x^{2}+(y-\displaystyle \frac{2}{3})^{2}=\frac{4}{9}$
The radius is $\displaystyle \frac{2}{3}$, and the center is at $(0,\displaystyle \frac{2}{3})$.
In polar coordinates, the center lies at $r_{0}=\displaystyle \frac{2}{3}, \theta_{0}=\pi/2,\qquad P(\frac{2}{3},\pi/2)$
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
$r=2(\displaystyle \frac{2}{3})\cos(\theta-\frac{\pi}{2})$
or
$r=\displaystyle \frac{4}{3}\cos(\theta-\frac{\pi}{2})$
using the identity
$\displaystyle \cos(\theta\pm\frac{\pi}{2})=\mp\sin\theta$,
$r=\displaystyle \frac{4}{3}\sin\theta$
