<s>[INST] Converting $x^2 + 6y - 9 = 0$ to polar. So far I got here
\begin{align}
(r\cos\phi)^2 & + 6 r \sin\phi- 9 = 0\\
(r\cos\phi)^2 & = 9 - 6r \sin\phi
\end{align}
 [/INST] $$x^2+6y=9$$
As always we choose $x=r \cos \theta$ and $y=r \sin \theta$
Thus $$(r \cos \theta)^2 +6r \sin \theta -9=0$$
With the identity $\cos^2 \theta = 1- \sin^2 \theta$ we find
$$r^2 - (r \sin \theta)^2 + 6r \sin \theta -9=0$$
Which equals $$(r \sin \theta -3)^2=r^2$$
Thus $r=r \sin \theta -3$ or $r= 3-r \sin \theta$ and subsequently we see $$r=\frac{3}{\sin \theta - 1} \lor r=\frac{3}{1+ \sin \theta} $$
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