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: 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 $(a,b)$. 
A: 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.
A: 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.
A: 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.
A: Although it's already been answered, it seemed like a fun way to procrastinate on my homework.
$(x-a)(x-a) + (y-b)(y-b) = aa + bb$
Substitute $x = rcos(θ)$ and $y = rsin(θ)$.
$(rcos(θ) - a)(rcos(θ) - a) + (rsin(θ) - b)(rsin(θ) - b) = aa + bb$
Multiply using FOIL.
$rrcos(θ)^2 - 2arcos(θ) + aa + rrsin(θ)^2 - 2brsin(θ) + bb = aa + bb$
Rearrange the terms.
$rrcos(θ)^2 + rrsin(θ)^2 -2arcos(θ) - 2brsin(θ) + aa + bb = aa + bb$
Factor out $r^2$.  Subtract $a^2$ from both sides. Subtract $b^2$ from both sides:
$rr(cos(θ)^2 + sin(θ)^2) -2arcos(θ) - 2brsin(θ) = 0$
Simplify left side by using the trig identity $sin(θ)^2 + cos(θ)^2 = 1$.
$rr(1) - 2arcos(θ) - 2brsin(θ) = 0$
Add $2arcos(θ)$ and $2brsin(θ)$ to both sides.
$rr(1) = 2arcos(θ) + 2brsin(θ)$
Simplify the left side.  Factor $r$ out of the right side.
$rr = r(2acos(θ) + 2bsin(θ))$
Divide both sides by $r$.
$(rr)/r = 2(acos(θ) + bsin(θ))$
Simplify the left side.
$r = 2acos(θ) + 2bsin(θ)$
