Family of circles orthogonal to two given circles

Question

The problem I am working on requires me to find the family of circles orthogonal to two given circles in the complex plane, namely $\delta \Delta(1)$ and $\delta \Delta (1,2)$ (which are the circles centered at origin with radius 1 and the circle centered at (1,0) with radius 2).

What I have tried to do till now is to send these circles to parallel lines using the transformation z $\rightarrow$ $1/(z+1)$. The images of these circles under this transformation are the vertical lines z = 1/2 and z =1 respectively. So the lines perpendicular to these would be the vertical lines and the preimages of these would give us the family of circles perpendicular to our given circles (as angle would be preserved).

However, I am unable find the preimages of these lines under the given map.

Any help would be greatly appreciated.

Answer 1

You are using a Mobius transformation. Its inverse is $$ \frac{-z+1}{z} $$

Note that a line is mapped to a circle or a line under a Mobius transformation.

Answer 2

You're almost there. $\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}$

Applying the map $\frac1{z+1}$ to $C((0,0),1)$ gives the line $\Re(z)=\frac12$, and applying this map to $C((1,0),2)$ gives the line $\Re(z)=\frac14$. As you state, the circles orthogonal to both of these are mapped to the horizontal lines $\Im(z)=t$.

The inverse function to $\frac1{z+1}$ is $\frac1z-1$. Applying this to the line $\Im(z)=t$ gives a circle with $\infty\to-1$ and $it\to-1-\frac it$ on its diameter. That is, $$ C\!\left(\left(-1,-\frac1{2t}\right),\frac1{2|t|}\right) $$ These are the circles tangent to the real axis at $(-1,0)$