To map
$$ (x-h)^2+(y-k)^2 = r^2 $$ to $$ (x-H)^2+(y-K)^2 = R^2 $$
in the complex plane is there a easy/quick way to determine $ a,b,c,d $ in terms of $ A,B,C,D $
the Moebius transformation $ w =\dfrac{az+b}{cz+d} \,? $
EDIT1:
(Wlog if take $d=1)$ I hope to see column matrix /vector $ f,f^{-1} $ algebraically expressed:
$$ (H,K,R)= f(a,b,c) (h,k,r) \, ; \,(h,k,r)\,=f^{-1}(a,b,c)(H,K,R)$$
I will solve an example hope it will help you to understand what Mobius transform do.
If we want to map the unit circle $x^2 + y^2 =1$ which can be written as $\mid z \mid = 1 $ into the circle $O: (x-2)^2 + (y-2)^2 = 4$ the circle with radius $2$ and center $(2,2)$ using Mobius maps. We can find many maps can do that.
To find one we will choose any three points choose any three points on the unit circle say $1, i , -1$ and any three points on the $O$ say $2 , 2i , 2+4i$. Our map will take ( I choose this ) $$1 \longmapsto 2 ,\;\;\;\ \ i\longmapsto 2i,\;\;\;\; -1 \longmapsto 2+4i$$
I will use the cross ratio (http://www.math.bas.bg/~rkovach/lectures/Moebius.pdf) $$(z,1,i,-1) = (w,2,2i,2+4i)$$ $$\frac{(z-1)(i--1)}{(z--1)(i-1)}= \frac{(w-2)(2i-(2+4i))}{(w-(2+4i))(2i-2)}$$ $$\frac{(z-1)(i+1)}{(z+1)(i-1)} = \frac{(w-2)(-2)(i+1)}{(w-2-4i)(2)(i-1)}$$ Solve it for $w$ $$w = 2 +2i -\frac{2i}{z}= \frac{(2+2i)z-2i}{z}$$ $a=2+2i , b = -2i , c = 1 , d = 0$
Mobius transform maps circles and lines into circles and lines and the map determined by three points.
The transform $$w= \frac{R}{r}\left(z-(h+i k)\right)+H+iK$$ Maps $(h, k-r)\longmapsto (H, K-R), \; \; (h, k+r)\longmapsto (H, K+R), \; \; (h+r, k)\longmapsto (H+R, K)$