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? 

I find it easier to go from polar to rectangular. 
$r=2a\cos\theta+2b\sin\theta$; $r^2=2ar\cos\theta+2br\sin\theta$; $x^2+y^2=2ax+2by$; $x^2-2ax+y^2-2by=0$; $(x-a)^2+(y-b)^2=a^2+b^2$; voila, circle of radius $\sqrt{a^2+b^2}$ centered at $(a,b)$. 
