<s>[INST] How to derive the polar equation of a circle. 
I have learnt that the standard equation of a circle whose centre is at (a,b) is
$$(x-a)^2 + (y-b)^2 = c^2
$$
I am trying to derive the polar equation of this circle but I am unfortunately stuck.
$$\begin{align}(x-a)^2+(y-b)^2&=c^2\\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=c^2\\
r^2\cos^2\theta-2ra\cos\theta+a^2+r^2\sin^2\theta-2rb\sin\theta+b^2&=c^2\\
r(r-2a\cos\theta-2b\sin\theta)&=c^2-a^2-b^2\end{align}$$
The following is stated in James Stewart's book on Precalculus but there is no proof stated in this book. So I am trying to reach a similar result by beginning at the standard equation of a circle.

 [/INST] You got off to a good start: Just plug in $x = r \cos\theta$ and $y = r \sin\theta$ in the standard rectangular-coordinate equation $(x-a)^2 + (y-b)^2 = c^2$
$$(r \cos\theta-a)^2 + (r \sin\theta-b)^2 = c^2$$
Or, if you need the equation expressed as a polynomial of $r$,
$$r^2 \cos^2\theta - 2ar\cos\theta + a^2 + r^2\sin^2\theta - 2br\sin\theta + b^2 = c^2$$
$$r^2(\cos^2\theta + \sin^2\theta) - 2r(a\cos\theta + b\sin\theta) + a^2 + b^2 - c^2 = 0$$
$$r^2 - 2(a\cos\theta + b\sin\theta)r + (a^2 + b^2 - c^2) = 0$$
Or, if you need to have an explicit formula for $r$ in terms of $\theta$, just use the Quadratic Formula,
$$r = \frac{2(a\cos\theta + b\sin\theta) \pm \sqrt{4(a\cos\theta + b\sin\theta)^2 - 4(a^2 + b^2 - c^2)}}{2}$$
$$r = \frac{2a\cos\theta + 2b\sin\theta \pm \sqrt{4a^2\cos^2\theta + 8ab\cos\theta\sin\theta+4b^2\sin^2\theta - 4a^2 - 4b^2 + 4c^2}}{2}$$
$$r = \frac{2a\cos\theta + 2b\sin\theta \pm 2 \sqrt{a^2(\cos^2\theta - 1) + 2ab\cos\theta\sin\theta+b^2(\sin^2\theta - 1) + c^2}}{2}$$
$$r = a\cos\theta + b\sin\theta \pm \sqrt{a^2(-\sin^2\theta) + 2ab\cos\theta\sin\theta+b^2(-\cos^2\theta) + c^2}$$
$$r = a\cos\theta + b\sin\theta \pm \sqrt{c^2 - (a\sin\theta - b\cos\theta)^2}$$
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