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 agree with the answer given by Gerry Myerson. The only detail left out is, is to complete the square on the terms: $x^2-2ax = (x-a)^2 - a^2$ and the same for: $y^2-2by = (y-b)^2 - b^2$
adding these 2 equations and equal to zero then gives the equation in Cartesian form as shown above by Gerry Myerson. I am just adding in the details for those that may not realize.
