<s>[INST] Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) =  a^2 + b^2$$
$$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$
Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought? 
 [/INST] I think your substitutions from the first line to the second aren't quite right.  It looks like you used $x=a\cos\theta$ and $y=b\sin\theta$, but you probably wanted $$\begin{align}x&=r\cos\theta\\y&=r\sin\theta.\end{align}$$
Using those, $$\begin{align}
(x-a)^2 + (y - b)^2 &= a^2 + b^2 \\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=a^2+b^2 \\
r^2\cos^2\theta+a^2-2ar\cos\theta+r^2\sin^2\theta+b^2-2br\sin\theta&=a^2+b^2 \\
r^2(\sin^2\theta+\cos^2\theta)+(a^2+b^2)-2r(a\cos\theta+b\sin\theta)&=a^2+b^2 \\
r^2-2r(a\cos\theta+b\sin\theta)&=0 \\
r^2=2r(a\cos\theta+b\sin\theta) \\
r=0\;\;\text{  or  }\;\;r=2a\cos\theta+2b\sin\theta
\end{align}$$
and since $r=0$ describes only the pole, which is also contained in the graph of $r=2a\cos\theta+2b\sin\theta$, $$r=2a\cos\theta+2b\sin\theta$$ is sufficient to describe the circle.
</s>