For $\alpha, \gamma \in \mathbb{R}, \alpha \neq 0, \beta \in \mathbb{C}$ and $\beta \bar{\beta} \geqslant \alpha \gamma$ the equation $\alpha z \bar{z}-\beta \bar{z}-\bar{\beta} z+\gamma=0$ describes a circle $C_{\alpha \beta \gamma}$ in the complex plane. Find its centre and radius. What does the equation describe if $\beta \bar{\beta}<\alpha \gamma$ ? Sketch the circles $C_{\alpha \beta \gamma}$ for $\beta=\gamma=1$ and $\alpha=-2,-1,-\frac{1}{2}, \frac{1}{2}, 1$.

Show that the complex function $f(z)=\beta \bar{z} / \bar{\beta}$ for $\beta \neq 0$ satisfies $f\left(C_{\alpha \beta \gamma}\right)=C_{\alpha \beta \gamma}$.

[Hint: $f(C)=C$ means that $f(z) \in C \forall z \in C$ and $\forall w \in C \quad \exists z \in C$ such that $f(z)=w .]$

For two circles $C_{1}$ and $C_{2}$ a function $m\left(C_{1}, C_{2}\right)$ is defined by

$$m\left(C_{1}, C_{2}\right)=\max _{z \in C_{1}, w \in C_{2}}|z-w|$$

Prove that $m\left(C_{1}, C_{2}\right) \leqslant m\left(C_{1}, C_{3}\right)+m\left(C_{2}, C_{3}\right)$. Show that

$$m\left(C_{\alpha_{1} \beta_{1} \gamma_{1}}, C_{\alpha_{2} \beta_{2} \gamma_{2}}\right)=\frac{\left|\alpha_{1} \beta_{2}-\alpha_{2} \beta_{1}\right|}{\left|\alpha_{1} \alpha_{2}\right|}+\frac{\sqrt{\beta_{1} \overline{\beta_{1}}-\alpha_{1} \gamma_{1}}}{\left|\alpha_{1}\right|}+\frac{\sqrt{\beta_{2} \overline{\beta_{2}}-\alpha_{2} \gamma_{2}}}{\left|\alpha_{2}\right|}$$