Question

What is the Cartesian form of `r^2-4r = sin(theta) - 5 cos(theta)`?

Answer 1

`4x^2+4y^2-5x+y=(x^2+y^2)^(3/2)`

Explanation:

To convert a polar coordinate `(r,theta)` in Cartesian form we use relation `r=sqrt(x^2+y^2)`, `rcostheta=x`, `rsintheta=y` and `theta=tan^(-1)(y/x)`. Using these relations

`r^2−4r=sintheta−5costheta` cann be written as

`r^3-4r^2=rsintheta−5rcostheta` or

`(x^2+y^2)^(3/2)-4(x^2+y^2)=y-5x` or

`4x^2+4y^2-5x+y=(x^2+y^2)^(3/2)`