Can someone explain how the cartesian equation is formed? Can someone explain how the cartesian equation of $r = 1 - \cos (\theta)$  is $x^4 + y^4 + 2x^2y^2 + 2x^3 + 2xy^2 - y^ 2 = 0$ ?
 A: 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.
A: $r^2=r-x\implies x^2+y^2+x=(x^2+y^2)^{\frac12}$. Then square both sides and arrange.
A: It is slightly tricky. Multiply through by $r$ and put $r\cos\theta=x$, that gives $r^2=r-x$ or $x=r-r^2$. Squaring and rearranging gives $2r^3=r^4+y^2$. Squaring again gives $4r^6=r^8+2y^2r^4+y^4$. Substituting $r^2=x^2+y^2$ gives a degree 8 polynomial which factors: $(x^4+y^4+2x^2y^2+2x^3+2xy^2-y^2)(x^4+y^4+2x^2y^2-2x^3-2xy^2-y^2)=0$.
So the first question is whether the second factor is valid or whether it is an artefact of all that squaring. Try putting $y=0,x=2$. That fails to satisfy the first factor, but does satisfy the second. It also fails to satisfy the original equation (whereas $y=0,x=-2$ satisfies the first factor and the original equation but not the second factor).
Since the original equation obviously generates a continuous curve, we conclude that we have $x^4+y^4+2x^2y^2+2x^3+2xy^2-y^2=0$ as required.
A: $$ r=\sqrt{x^2+y^2} , \, \cos \theta = \frac {x} { \sqrt{x^2+y^2}  }$$
Plug them in, squaring it at least twice to get rid of radicals.
