# How do you convert x^2+4x+y^2+4y=0 to polar form?

Oct 19, 2016

$r + 4 \sqrt{2} \cos \left(\theta - \frac{\pi}{4}\right) = 0$

#### Explanation:

The relation between polar coordinates $\left(r , \theta\right)$ and rectangular Cartesian coordinates $\left(x , y\right)$ are given by

$x = r \cos \theta$ and $y = r \sin \theta$ or ${r}^{2} = {x}^{2} + {y}^{2}$

Hence ${x}^{2} + 4 x + {y}^{2} + 4 y = 0$ can be written as

${x}^{2} + {y}^{2} + 4 x + 4 y = 0$

or ${r}^{2} + 4 r \cos \theta + 4 r \sin \theta = 0$

or ${r}^{2} + 4 r \left(\cos \theta + \sin \theta\right) = 0$

or $r + 4 \left(\cos \theta + \sin \theta\right) = 0$

or $r + 4 \sqrt{2} \left(\cos \theta \times \frac{1}{\sqrt{2}} + \sin \theta \times \frac{1}{\sqrt{2}}\right) = 0$

or $r + 4 \sqrt{2} \left(\cos \theta \cos \left(\frac{\pi}{4}\right) + \sin \theta \sin \left(\frac{\pi}{4}\right)\right) = 0$

or $r + 4 \sqrt{2} \cos \left(\theta - \frac{\pi}{4}\right) = 0$