<s>[INST] Deriving polar graphs There is just some confusion I have with polar graphs, for instance there are some well known general forms of the polar graphs such as the circle, Limaçon,  rose, Lemniscate.
The general equation for a circle with radius $\frac{a}{2}$ is given by either:
$$r = a \cos\theta \space \space \text{or}\space \space r = a \sin \theta$$
Then for Cardioids or Limacons are:
$$r = a \pm b\cos\theta \space \space \text{or}\space \space r = a \pm b \sin \theta$$
For Roses, we have:
$$r = a \cos n\theta \space \space \text{or}\space \space r = a\sin n\theta$$
Finally for Lemniscates we have:
$$r^2 = a^2 \cos n\theta \space \space \text{or}\space \space r^2 = a^2 \sin n\theta$$
So on and so forth. However what I am truly confused about is how does on come to derive that these shapes are indeed these algebraically? For instance I think I can derive the circle equation using $x$, $y$ for instance:
$$\bigg(x-\frac{a}{2}\bigg)^2+y^2 = \bigg(\frac{a}{2}\bigg)^2$$
$$x^2+y^2 = ax$$
$$r^2 = ar\cos\theta$$
$$r = a\cos\theta$$
Hence the above polar equation has the circle equation with center $(\frac{a}{2},0)$
So My QUESTION is how can we derive the other $3$ special cases for polar graphs starting for our $x$ and $y$ like we have for the circle case? 
 [/INST] Degree 1 polynomials describe lines
Degree 2 polynomials describe lines the conic sections, parabola, circle, ellipse, hyperbola.
Degree 3 polynomials describe "elliptic curves"
You will need a $4^{th}$ degree$^+$ polynomial.
A cartiod
$r = a(1+\cos \theta)\\
r^2 = a(r+r\cos \theta)\\
x = r\cos \theta, r = \sqrt {x^2 + y^2}\\
x^2 + y^2 - ax = a\sqrt {x^2 + y^2}\\
(x^2 + y^2 - ax)^2 = a^2(x^2 + y^2)$
The rest can be found using similar approaches.
4-petaled rose.
$r = \cos 2\theta\\
r = \cos^2\theta - \sin^2\theta\\
r^3 = (r\cos\theta)^2 - (r\sin\theta)^2\\
(x^2 + y^2)^\frac 32 = x^2 - y^2\\
(x^2 + y^2)^3 = (x^2 - y^2)^2$
3 petaled rose
$r = \cos 3\theta\\
r = \cos^3 \theta - 3\cos\theta\sin^2\theta\\
r^4 = r^3\cos^3 \theta - 3r\cos\theta r^2\sin^2\theta\\
(x^2+y^2)^2 = x^3 - 3xy^2 $
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