I know this might be an unusual question, but please bear with me.
In the book ¨An Introduction to Maximum Principles and Symmetry in Elliptic Problems¨ There is the following example of transformation:
right afterwords it is said that the ball $B(p , a)$ has the image $B_*=B(p_* , a_*)$ where $ p_* = \frac{1}{1-(a/2b)²}p $ and $ a_* = \frac{1}{1-(a/2b)²}a $ What is he talking about? $p$ is a fixed point? where do we map $p$ to $p_*$?
thanks very much.
Yes, by design. Since its distance from $0$ is $2b$, we have $\dfrac{4b^2}{r^2}p=p$. More intuitively, all points lying on a sphere are fixed by the inversion in that sphere.
We don't. The ball $B(p,a)$ is mapped onto $B(p_*,a_*)$, but the old center does not go to the new center. This is just how Möbius transformations work. As a simple 1D example, consider the transformation $y=1/x$ which maps the "ball" (line segment) $[2,6]$ onto $[1/6,1/2]$; the old center $(2+6)/2=4$ is mapped to $1/4$, which is different from the new center $(1/6+1/2)/2 = 1/3$.
The typical way one figures the center and radius of the image of a ball is to find the image of two antipodal points $x,y$ on the original ball. The midpoint of the images is the new center, and their half-distance is the new radius.