Question

How do you convert `r=6cosθ` into a cartesian equation?

Answer 1

I found the equation:`x^2+y^2-6x=0` which represents a circle.

Explanation:

Consider the connection between Cartesian and Polar:
`r=sqrt(x^2+y^2)`
`theta=arcrtan(y/x)`

And:

`x=rcos(theta)`
`y=rsin(theta)`

In your case you get:
`sqrt(x^2+y^2)=6x/r`
`sqrt(x^2+y^2)=6x/sqrt(x^2+y^2)`
`x^2+y^2-6x=0`