Answer
$y^2-16-8x=0$
Work Step by Step
After multiplying with the denominator of the fraction: $r(1-\cos{\theta})=4\\r-r\cos{\theta}=4\\r=4+r\cos{\theta}\\r^2=(4+r\cos{\theta})^2$.
We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$.
Hence $r^2=(4+r\cos{\theta})^2$ becomes: $x^2+y^2=(4+x)^2\\x^2+y^2=16+8x+x^2\\y^2-16-8x=0$
