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? 

A 'backwards' answer:
The general polar equation of a circle of radius $\rho$ centered at $(r_0,\theta_0)$ is 
$$r^2-2 r r_0 \cos(\theta-\theta_0) + r_0^2 = \rho^2.$$
When $r_0 = \rho$, this reduces to (ignoring the $r=0$ solution)
$$r = 2(r_0 \cos \theta_0) \cos \theta + 2(r_0 \sin \theta_0) \sin \theta,$$
choosing $(r_0,\theta_0)$ such that $a = r_0 \cos \theta_0$, $b = r_0 \sin \theta_0$ gives the required form.
