# How do you convert r = 4 cosθ + 4 sinθ into a cartesian equation?

Apr 18, 2018

${x}^{2} + {y}^{2} - 4 \left(x + y\right) = 0$

#### Explanation:

We know that $x = r \cos \theta$ and $y = r \sin \theta$

$\Rightarrow \frac{x}{r} = \cos \theta$ and $\frac{y}{r} = \sin \theta$

Put the above values in the equation :-

$r = 4 \cos \theta + 4 \sin \theta$

$\Rightarrow r = \frac{4 \left(x + y\right)}{r}$

$\Rightarrow {r}^{2} = 4 \left(x + y\right)$

Also we know that ${x}^{2} + {y}^{2} = {r}^{2}$ ; put in the equation we get :-

$\Rightarrow {x}^{2} + {y}^{2} = 4 \left(x + y\right)$

$\therefore {x}^{2} + {y}^{2} - 4 \left(x + y\right) = 0$ which is the cartesian form of the given equation.