By the [Möbius Decomposition](/theorems/810) theorem, it suffices to check the three elementary types. A circle in $\mathbb{C}_\infty$ is defined by $Az\bar{z} + Bz + \bar{B}\bar{z} + C = 0$ with $A, C \in \mathbb{R}$ and $|B|^2 > AC$. **Step 1: Dilation $f(z) = az$ with $a \neq 0$.** Substituting $z = w/a$ (where $w = f(z)$) into the circle equation gives: \begin{align*} A \frac{w\bar{w}}{|a|^2} + B\frac{w}{a} + \bar{B}\frac{\bar{w}}{\bar{a}} + C = 0, \end{align*} which is the circle $S_{A/|a|^2,\; B/a,\; C}$. The condition $|B/a|^2 > (A/|a|^2) \cdot C$ is equivalent to $|B|^2 > AC$, so this is again a circle. **Step 2: Translation $f(z) = z + b$.** Substituting $z = w - b$ gives: \begin{align*} A(w-b)\overline{(w-b)} + B(w-b) + \bar{B}(\bar{w} - \bar{b}) + C = 0, \end{align*} which rearranges to $Aw\bar{w} + (B - A\bar{b})w + (\bar{B} - Ab)\bar{w} + (A|b|^2 - B\bar{b} - \bar{B}b + C) = 0$. This has the form $A'w\bar{w} + B'w + \bar{B'}\bar{w} + C' = 0$ with $A' = A \in \mathbb{R}$ and $C' \in \mathbb{R}$. One verifies $|B'|^2 > A'C'$, so this is a circle. **Step 3: Inversion $f(z) = 1/z$.** Substituting $z = 1/w$ into $Az\bar{z} + Bz + \bar{B}\bar{z} + C = 0$ gives: \begin{align*} \frac{A}{w\bar{w}} + \frac{B}{w} + \frac{\bar{B}}{\bar{w}} + C = 0. \end{align*} Multiplying through by $w\bar{w}$: $A + B\bar{w} + \bar{B}w + Cw\bar{w} = 0$, which is the circle $S_{C,\;\bar{B},\;A}$. The condition $|\bar{B}|^2 > CA$ is the same as $|B|^2 > AC$, confirming it is a circle. **Step 4: Conclude.** Since each elementary type maps circles to circles, and every Möbius map is a composition of these types, every Möbius map preserves circles.