 Solution

A Möbius transformation is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto 1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equationwhere $\alpha ,\gamma \in \mathbb{R}$; the case $\alpha =0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w{|}^2$ to obtainwhich is again such an equation. The generators, and hence every Möbius transformation, therefore preserve Generalized circle under a Möbius transformation.

Now letIf $c\ne 0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the subgroup in question is the Affine subgroup of the Möbius group, consisting of $z\mapsto az+b$ with $a\ne 0$.

It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne 0$, then $T_b$ is affine buthas a finite pole and is not affine. Hence conjugation by $I$ does not preserve the subgroup.