<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] By to reach a similar result I understood that you are interested in an unparametrized form.
We obtain polar equation by continuing the derivation to its  logical conclusion.
$$ r(r-2a\cos\theta-2b\sin\theta)=c^2-a^2-b^2$$
Distance to center from origin   $OC=\sqrt {a^2+b^2}$ and power of circle /squared tangent length
$$ a^2+b^2-c^2= OT^2=T^2 $$
$$ = r(r- 2a \cos\theta-2b \sin\theta)=OT^2$$
$$ = r(r- 2  \sqrt {a^2+b^2}\cdot ( \frac{a}{\sqrt {a^2+b^2}}\cos \theta +  \frac{b}{\sqrt {a^2+b^2}}\sin \theta) )$$
$$ \alpha=\angle COx= \tan^{-1}\frac{b}{a}, \;\theta=\angle AOx\;$$

$$ =r(r- 2  \sqrt {a^2+b^2}( \cos\alpha \cos \theta +  \sin \alpha\sin \theta) )$$
$$OT^2= r\left(r- 2  \sqrt {a^2+b^2} \cos(\alpha- \theta)\right) \tag 1 $$
This can be readily recognized as the property of circle with a constant product of segments made by intersecting chords. Drop perpendicular CM onto OA. M is mid-point of AB, OMC is a right triangle
$$OM=\frac{OA+OB}{2}= OC \cos (\alpha-\theta)$$
$$ OA=r, OB=r- 2\sqrt{a^2+b^2}\cos (\alpha-\theta)= r- 2 Q $$
$$Q=\sqrt{a^2+b^2}\cos (\alpha-\theta) $$
$$ OA\cdot OB=T^2 \;, r(r-2Q)=OT^2;\;\tag 2 $$
Solving the quadratic we obtain
$$ r= Q\pm\sqrt{Q^2-T^2} $$
The radius vector intersects a circle at two points $(A,B)$. Finally the required polar coordinate equations are
$$ r= \sqrt{a^2+b^2}\cos (\alpha-\theta) \pm \sqrt{c^2-(a^2+b^2)\sin^2 (\theta-\alpha)} \tag 3 $$
Positive sign for outer part, negative sign for inner part and radicals disappear at tangent point $r=T.$
It can be verified by trigonometry for any Circle radius $c$ centered at $(a,b)$  the radius  must always be  $r=T $ at point of tangency.I.e.,
$$ @\; \theta = \tan^{-1}\frac{b}{a} \pm \sin^{-1}\frac{c}{\sqrt{a^2+b^2}}, \; r=T\; \text{ always}$$
Extreme radii $ \; @ \theta= \alpha\; $ are
$$ \sqrt{a^2+b^2}\pm c $$
All the details are shown in the graph drawn for assumed constants
$$(a,b,c)=(4,3,2), r_{max}=7, r_{min}=3, T=\sqrt{21}\approx 4.5826 $$
A Cartesian $(r-\theta)$ plot is also given for a distorted circle in an alternate coordinate representation, as we are more familiar using Cartesian coordinates. It may be ignored at first reading, it is intended to identify four cardinal points.. ( two extreme radii, two points of tangency).
As an exercise you can make the necessary change for a negative power eccentric circle.
Actually I hope it is not the " general" polar equation given in the text-book. It pertains only to two particular cases when the circles pass through the origin, when centered on $(x,y)$ axes, $r_{min}=0 $ but is not about general eccentric circles placed in arbitrary quadrants.
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