## Calculus (3rd Edition)

$9\left( {{x^2} + {y^2}} \right) = {\left( {{x^2} + {y^2} - 2y} \right)^2}$ as an equation in polar coordinates: $r = 3 + 2\sin \theta$
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 $9{r^2} = {\left( {{r^2} - 2r\sin \theta } \right)^2}$ $3r = {r^2} - 2r\sin \theta$ $3 = r - 2\sin \theta$ $r = 3 + 2\sin \theta$