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

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Answer 1 · 2016-07-30T17:51:57

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

The relation between Cartesian coordinates $(x.y)$ and $(r,theta)$ is given by $x=rcostheta$ and $y=rsintheta$ and $r^2=x^2+y^2$

Hence, $r(2-costheta)=2$ can be written as

$2r-rcostheta=2$ or

$2sqrt(x^2+y^2)-x=2$ or

$2sqrt(x^2+y^2)=2+x$ and squaring

$4(x^2+y^2)=(2+x)^2=4+4x+x^2$ or

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

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

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