Question

How do you convert `(r-1)^2= - sin theta costheta +costheta` to Cartesian form?

Answer 1

`x^2+y^2-2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)+x/sqrt(x^2+y^2)`

Explanation:

The relation between polar coordinates `(r,theta)` and Cartesian coordinates `(x,y)` is given by

`x=rcostheta`, `y=rsintheta` and `r^2=x^2+y^2`

Hence, `(r-1)^2=-sinthetacostheta+costheta`

`hArrr^2-2r+1=-sinthetacostheta+costheta`

or `x^2+y^2-2sqrt(x^2+y^2)+1=-(y/r*x/r)+x/r`

or `x^2+y^2-2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)+x/sqrt(x^2+y^2)`