Is it possible to convert the polar equation $\ r = k \cos (\theta n) + 2$ into cartesian form?

Question

Is it possible to convert the polaer equation $$\ r = k \cos (\theta n) + 2$$ into cartesian form? Here, $k$ is some constant and $n$ is any positive whole number greater than $2$.

The farthest that I managed to get was:

$$r^2 = kr \cos(\theta n) +2r$$

$$\to\qquad r^2 - 2r - kr\cos(\theta n) = 0$$

$$\to \qquad x^2 +y^2 - 2 \sqrt{x^2 +y^2} -kr \cos (\theta n) = 0$$

I noticed that $\cos(\theta n)$ is reminiscent of the Chebychev polynomials and figured that $r\cos(\theta n)$ could also be generalized into polyomials in terms of $x$, but that turns out to be impossible.

Answer 1

If you want an expression that holds for all $n \geq 2$ then it might be a tad ugly in Cartesian, because:

$$\cos(n\theta) = \sum_{\text{even }k} (-1)^{k/2}{n \choose k}\cos^{n-k} \theta \sin^k \theta \\ = (x^2+y^2)^{-n/2}\sum_{\text{even }k} (-1)^{k/2}{n \choose k} x^{n-k}y^k.$$

But this will get you there.

Answer 2

If $T_n$ is the $n$-th Chebyshev polynomial of the first kind then you have $T_n(\cos \theta) = \cos n \theta$ and so your equation can be written as

$$ \sqrt{x^2 + y^2} = k T_n \left( \frac{x}{\sqrt{x^2 + y^2}} \right) + 2. $$

If we multiply both sides by $(x^2 + y^2)^{\frac{n}{2}}$, we can rewrite the expression above as $f(x,y) = 0$ for a function $f$ which is well-defined for all $(x,y) \in \mathbb{R}^2$. The level set $f(x,y) = 0$ will be the same as the level set of your equation except that the origin $(0,0)$ will always belong to $f(x,y) = 0$ while the origin might not lie on your polar equation.