# What is the polar form of (16,-4?

Nov 30, 2017

$\left(4 \sqrt{17} , - 0 , 245\right)$

#### Explanation:

The quickest method is to use the formula:

$\left(x , y\right) = \left(\sqrt{{x}^{2} + {y}^{2}} , \arctan \left(\frac{y}{x}\right)\right)$

Below is an alternative method.

$x = r \cos \theta$

$y = r \sin \theta$

$\therefore$

$16 = r \cos \theta$ [ 1 ]

$- 4 = r \sin \theta$ [ 2 ]

Squaring both equations:

$256 = {r}^{2} {\cos}^{2} \left(\theta\right)$

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

$272 = {r}^{2} {\cos}^{2} \left(\theta\right) + {r}^{2} {\sin}^{2} \left(\theta\right)$

$272 = {r}^{2} \left({\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right)$

$272 = {r}^{2} \left(1\right) \implies r = \pm \sqrt{272} = \pm 4 \sqrt{17}$

We only need the positive root, negative radii can exist, and a point can be represented by many different polar coordinates, unlike Cartesian coordinates which are unique.

$\theta = \arctan \left(\frac{y}{x}\right) = \arctan \left(- \frac{4}{16}\right) = \arctan \left(- \frac{1}{4}\right) = - 0.24498$

Polar coordinate:

$\left(4 \sqrt{17} , - 0 , 245\right)$