<s>[INST] Find the Cartesian equation corresponding to $r = \frac{5}{3-2\cos(\theta)}$ 
Find the Cartesian equation corresponding to $r =
 \frac{5}{3-2\cos(\theta)}$

I got it into the form:
$(x^2 + y^2)(3-2x)^2 = 25$
and can see that maybe the equation of a circle will appear, but I can't seem to get much further without complicating things. I have a note saying that completing the square was used but I can't see how or where.
 [/INST] Rearranging we get that $3r - 2r\cos(\theta) = 5$. Note that $r = \sqrt{x^2 + y^2}$ and $r \cos(\theta) = x$.
This gives us $3 \sqrt{x^2+y^2} -2x = 5$.
Hence,
\begin{align}
9(x^2+y^2) & = (2x+5)^2\\
9x^2+9y^2 & = 4x^2 + 20x + 25\\
5x^2 -20x + 9y^2 & = 25\\
5(x-2)^2 + 9y^2 & = 45\\
\frac{(x-2)^2}{3^2} + \frac{y^2}{\left(\sqrt{5} \right)^2} & = 1
\end{align}
The above is an ellipse with center $(2,0)$ with semi-major axis along $x$ of length $3$ and semi-minor axis along $y$ of length $\sqrt{5}$.
EDIT
In general, $$r(\theta) = \dfrac{a(1-e^2)}{1 \pm e \cos(\theta)}$$ represents an ellipse whose semi-major axis is $a$ and eccentricity $e$.
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