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Complete the square in the real-valued complex plane equation:Dividing by $\alpha$ gives center and radiusIf the radicand is zero, the locus is the single point $c$, a degenerate circle. If it is negative, the locus is empty. The absolute value of $\alpha$ is essential even when $\alpha <0$.

For the requested sketch, the five centers and radii are respectively $(-1/2,\sqrt{3}/2)$, $(-1,\sqrt{2})$, $(-2,\sqrt{6})$, $(2,\sqrt{2})$ and $(1,0)$, all centers on the real axis.

The original PDF's map is $f(z)=(\beta /\bar{\beta })\bar{z}$; the converted TeX drops the conjugate on its denominator. Write $\beta =|\beta |e^{i\theta }$, so $f(z)=e^{2i\theta }\bar{z}$. This is the reflection symmetry of a complex circle equation. It fixes $c=\beta /\alpha$ and satisfiesThus it maps every point of the circle into the same circle. Conversely, for any point $w$ on that circle, $z=f(w)$ also lies on it and $f(z)=w$. This proves $f(C)=C$, including the surjectivity required by the question.

For any three nonempty circles, fix $u\in C_3$. The triangle inequality gives, for every $z\in C_1,w\in C_2$,Taking the maximum over $z,w$ proves the requested inequality.

For centers $c_1,c_2$ and radii $r_1,r_2$, the maximum separation of two circles is bounded above by $|c_1-c_2|+r_1+r_2$. If the centers differ, choose $z=c_1+r_{1}v$ and $w=c_2-r_{2}v$, where $v=(c_1-c_2)/|c_1-c_2|$, to attain the bound. If the centers coincide, any unit $v$ gives opposite points attaining $r_1+r_2$. ThereforeThe function $m$ is not a metric on positive-radius circles, since $m(C,C)=2r$, even though it satisfies this triangle-type inequality.