# How do you convert r(2 - cosx) = 2 into cartesian form?

Jul 30, 2016

$3 {x}^{2} + 4 {y}^{2} - 4 x - 4 = 0$

#### Explanation:

The relation between Cartesian coordinates $\left(x . y\right)$ and $\left(r , \theta\right)$ is given by $x = r \cos \theta$ and $y = r \sin \theta$ and ${r}^{2} = {x}^{2} + {y}^{2}$

Hence, $r \left(2 - \cos \theta\right) = 2$ can be written as

$2 r - r \cos \theta = 2$ or

$2 \sqrt{{x}^{2} + {y}^{2}} - x = 2$ or

$2 \sqrt{{x}^{2} + {y}^{2}} = 2 + x$ and squaring

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

$3 {x}^{2} + 4 {y}^{2} - 4 x - 4 = 0$

graph{3x^2+4y^2-4x-4=0 [-1.708, 3.292, -1.3, 1.5]}