Can someone explain how the cartesian equation is formed? Can someone explain how the cartesian equation of $r = 1 - \cos (\theta)$  is $x^4 + y^4 + 2x^2y^2 + 2x^3 + 2xy^2 - y^ 2 = 0$ ?

Since my comment was pseudo-copied to an answer (i.e. an answer identical to my comment was posted) I thought I'd just post a full solution completing that comment.


*

*Remember:
$$\left\{\begin{array}{c}
r\cos\theta=x \\
r\sin\theta=y \\
r=\sqrt{x^2+y^2} \\
\theta=\arctan\frac{y}{x}\text{ well, in some cases}
\end{array}\right.;$$

*Multiply by $r$ to get $r^2=r-r\cos\theta$;

*Substitute the above equations to get $x^2+y^2=\sqrt{x^2+y^2}-x$;

*Carry $x$ over to the LHS and square to get $x^2+y^2=(x^2+y^2+x)^2$;

*Expand the square on the right and get $x^4+y^4+x^2+2x^2y^2+2xy^2+2x^3=x^2+y^2$;

*$x^2$'s cancel out, $y^2$ goes to the left side, and we get:
$$x^4+y^4+2x^3-y^2+2x^2y^2+2xy^2=0.$$
Which is precisely what we wanted.
