# How do you convert x^2+(y-3)^2=9 to polar form?

Apr 17, 2017

#### Explanation:

Given: ${x}^{2} + {\left(y - 3\right)}^{2} = 9$

expand the square:

${x}^{2} + {y}^{2} - 6 y + 9 = 9$

Substitute ${r}^{2} \text{ for } {x}^{2} + {y}^{2}$:

${r}^{2} - 6 y = 0$

Substitute $r \sin \left(\theta\right)$ for y:

${r}^{2} - 6 r \sin \left(\theta\right) = 0$

We can discard a factor of r because the root $r = 0$ is duplicated at $\theta = 0$

$r - 6 \sin \left(\theta\right) = 0$

$r = 6 \sin \left(\theta\right)$