Conformal mapping of two circles into a line and a circle I have the following conformal mapping:

I need to find $\lambda = f(\zeta)$ and its reverse. Zeros on the figure are given, the axis are oriented as usual. The resulting distance between the line and the circle does not matter.
I know how to map the left arrangement to a circle and a line:
$$f(\zeta) = -i \frac{\zeta + iA}{\zeta - iA}$$
but how do I make the bottom circle transform into a circle with a unit radius?
Unfortunately it's not a homework so you may give an answer as is, should it be easier.
 A: We can think of the two circles in the Riemann Sphere opposite to each other. Then we need a scaling transformation to change the original circle into the equator, and the second circle must be still a circle. Now the equator of the Riemann Sphere is the $x$-axis in the complex plane. For two circles in the general position some rotation, etc should be required. It seems to me the situation is rigid enough that you cannot expect a transformation turning the second circle into unit radius.
A: It turned out really easy, to transform a line and a circle to a line and a unit circle in the arrangement shown above one only has to divide by the radius of the circle.
So the final transform is:
$$\lambda(\zeta) = -i \frac{\zeta + iA}{\zeta - iA} \frac{(H+A)^2-a^2}{2Aa}$$
Gnuplot for you to try ($\text{Eta} = \lambda$):
set terminal postscript enhanced
set output "Plot.ps"
set parametric

set samples 100
set isosamples 10

set nokey

set size ratio -1
set xrange [-3 : 3]
set yrange [-10 : 2]
set trange [0: 2*pi]

Zeta(t) = A*exp(t*{0,1})
Eta(t) = (-{0,1} * (Zeta(t) + {0,1} * A) / (Zeta(t) - {0,1} * A)) * ((H+A)**2-a**2)/(2*A*a)

zeta(t) = {0,-1} * H + a*exp(t*{0,1})
eta(t) = -{0,1} * (zeta(t) + {0,1} * A) / (zeta(t) - {0,1} * A) * ((H+A)**2-a**2)/(2*A*a)

A = 1
a = 0.2
H = 2
set multiplot
plot real(Zeta(t)),imag(Zeta(t)) lt 3 ,\
     real(Eta(t)),imag(Eta(t))   lt 1 ,\
     real(zeta(t)),imag(zeta(t)) lt 3 ,\
     real(eta(t)),imag(eta(t))   lt 1

