# How do you convert x² + (y + 4)² = 16 to polar form?

Oct 7, 2016

${r}^{2} + 8 r \sin \theta = 0$

#### Explanation:

The relation between polar coordinates $\left(r , \theta\right)$ and Cartesian coordinates $\left(x , y\right)$ is $x = r \cos \theta$ and $y = r \sin \theta$.

Therefore  can be written as

${\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta + 4\right)}^{2} = 16$

or ${r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta + 8 r \sin \theta + 16 = 16$

or ${r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + 8 r \sin \theta = 0$

or ${r}^{2} + 8 r \sin \theta = 0$