Question

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

Answer 1

We know the relations

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

where `(x,y) and (r,theta)` respectively represent the coordinates of same point in cartesian and in polar form.

Given equation

`(r-1)^2=-sinthetacostheta+cos^2theta`

`=>r^2(r-1)^2=-rsintheta* rcostheta+r^2cos^2theta`

`=>(r^2-r)^2=-rsintheta* rcostheta+r^2cos^2theta`

`=>(x^2+y^2-sqrt(x^2+y^2))^2=-y* x+x^2`

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

This equation represents the cartesian form of the given polar equation.

Answer 2

`(x^2+y^2)(x^2+y^2-sqrt(x^2+y^2))+y(x+y)=0`. Illustrative graph is inserted.

Explanation:

`(r-1)^2=r^2-2r+1=-sin theta cos theta + cos^2 theta`. So,

`r^2-2r=-sin theta cos theta -(1-cos^2theta)`

`=-sin theta cos theta -sin^2theta =-(y(x+y))/r^2`. using

`cos theta = x/r and sin theta = y/r`, where `r = sqrt(x^2+y^2)`

Cross multiplying and using `r=sqrt(x^2+y^2)`,

`(x^2+y^2)(x^2+y^2-2sqrt(x^2+y^2))+y(x+y)=0`

graph{(x^2+y^2)(x^2+y^2-2sqrt(x^2+y^2))+y(x+y)=0 [-5, 5, -2.5, 2.5]}