Prove that the image of the center of a given circle is never the same as that of the circle image under a linear fractional map.
Given some linear fractional map$f(z)=\frac{az+b}{cz+d}$, that is $ad-bc\neq0, c\neq0$ and a circle on $\Bbb C$ not passing through $z_0=\frac{-d}{c}$, so that the image is another circle, prove that the image of the center of the given circle is never the same as that of the image.
$a, b, c, d\in\Bbb R$ and $z\in\Bbb C$
I'm not sure how to generalize this proof for any circle with the given specifications and its image under the linear fractional map.
By composing $f$ with affine maps before and after, we can assume WLOG that the linear fractional map is $z \to 1/z$, and the centre of the first circle is at $p$ on the nonnegative real axis. The radius is $r>0$ where $r \ne p$. The image of $0$ is $\infty$, which is not the centre of a circle, so we can assume $p \ne 0$. The closest and farthest points to $0$ on the circle are at $p-r$ and $p+r$. Their images $1/(p-r)$ and $1/(p+r)$ are the farthest and closest points to $0$ on the image of the circle, so the centre of that image is at $$ q = \dfrac{1}{2(p-r)} + \dfrac{1}{2(p+r)}$$ Simplify $q - 1/p$ and you'll see that $q \ne 1/p$.