In regards to this question, when a circle is transformed into a circle under the map $w = 1/z$ the center is never mapped onto the center of the image circle. I am concerned about the alternative approach the person providing the answer mentions. They say that we can show this result by using $$ x = \frac{u}{u^{2} + v^{2}},\qquad y = -\frac{v}{u^{2} + v^{2}}. \quad{(1)} $$
I am able to get this substitution as follows from rewriting $z = 1/w$. However, to see an example of the claim, I consider the unit circle $|z| = x^2 + y^2 = 1.$ This has center at the origin. If I substitute the equations in (1) to this equation, I get $$x^2 + y^2 = \frac{1}{u^2 + v^2} = 1 \Rightarrow u^2 + v^2 = 1.$$ This still has center at the origin, which is not the claim. What am I doing wrong? Also, how could we generalize from this to show that the center is "never" mapped onto the center of the image circle using this approach?
As $0$ is mapped to $\infty$, we need only consider circles that are neither around the origin nor pass through the origin. If the radius is $r$ and the distance to the centre is $a$, the line through $0$ and the centre intersects the circle in two points at oriented distances $a+r$ and $a-r$ from the origin (and of course, the line contains the centre) After inversion, these three points have oriented distances $\frac1{a+r}$, $\frac1a$, $\frac1{a-r}$ from the origin. For the centre to map to the centre, we need $$ \frac1a=\frac12\left(\frac1{a+r}+\frac1{a-r}\right)=\frac{(a-r)+(a+r)}{2(a^2-r^2)}=\frac a{a^2-r^2}$$ and therefore $a^2=a^2-r^2$, i.e., $r=0$. Indeed such degenerate circles may be seen as an exception. Bu they are the only ones.