Answer
$9\left( {{x^2} + {y^2}} \right) = {\left( {{x^2} + {y^2} - 2y} \right)^2}$ as an equation in polar coordinates:
$r = 3 + 2\sin \theta $
Work Step by Step
Since $x = r\cos \theta $ and $y = r\sin \theta $, we get ${x^2} + {y^2} = {r^2}$. Substituting in this equation:
$9\left( {{x^2} + {y^2}} \right) = {\left( {{x^2} + {y^2} - 2y} \right)^2}$
we get
$9{r^2} = {\left( {{r^2} - 2r\sin \theta } \right)^2}$
$3r = {r^2} - 2r\sin \theta $
$3 = r - 2\sin \theta $
$r = 3 + 2\sin \theta $
