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

May 3, 2016

$r = 8 \sin \theta$

#### Explanation:

If $\left(r , \theta\right)$ is in polar form and $\left(x , y\right)$ in Cartesian form the relation between them is as follows:

$x = r \cos \theta$, $y = r \sin \theta$, ${r}^{2} = {x}^{2} + {y}^{2}$ and $\tan \theta = \frac{y}{x}$

Hence, ${x}^{2} + {\left(y - 4\right)}^{2} = 16$ can be written as

${x}^{2} + {y}^{2} - 8 y + 16 = 16$ or

${x}^{2} + {y}^{2} - 8 y = 0$ or

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

$r \left(r - 8 \sin \theta\right) = 0$ dividing by $r$

$r - 8 \sin \theta = 0$ or

$r = 8 \sin \theta$

May 3, 2016

$r = 8 \sin \theta$

#### Explanation:

To convert from Cartesian to Polar coordinates use the following formulae that link them.

• x=rcostheta" and " y=rsintheta

${x}^{2} + {y}^{2} - 8 y + 16 = 16 \text{ (expanding bracket) }$

$\Rightarrow {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta - 8 r \sin \theta + 16 - 16 = 0$

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

using the identity: ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

$\Rightarrow {r}^{2} = 8 r \sin \theta \text{ and dividing both sides by r }$

$\Rightarrow r = 8 \sin \theta$