Question

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

Answer 1

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

Explanation:

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.`