How do you convert $r = 2-2cos(theta)$ into rectangular form?

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Answer 1 · 2016-06-25T09:41:15

$(x^2+y^2)^2+4x(x^2+y^2)-4y^2=0.$

The conversion formula reqd. are $x=rcostheta, y=rsintheta, r=sqrt(x^2+y^2).$

Putting $costheta=x/r$ in the given eqn., we get, $r=2-2x/r,$ or, $r^2=2r-2x,$ i.e., $x^2+y^2=2sqrt(x^2+y^2)-2x rArr {(x^2+y^2)+2x)^2=4(x^2+y^2).$

$:. (x^2+y^2)^2+4x(x^2+y^2)+4x^2-4x^2-4y^2=0.$

$:. (x^2+y^2)^2+4x(x^2+y^2)-4y^2=0.$

How do you convert r = 2-2cos(theta) r = 2-2cos(theta) into rectangular form?