= Solution
Complete the square in the real-valued <complex plane> equation:
$$
\alpha\left|z-\frac\beta\alpha\right|^2=\frac{|\beta|^2-\alpha\gamma}{\alpha}.
$$
Dividing by $\alpha$ gives \b[center and radius]
$$
\boxed{c=\frac\beta\alpha,\qquad r=\frac{\sqrt{|\beta|^2-\alpha\gamma}}{|\alpha|}.}
$$
If 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 $(-1/2,\sqrt3/2)$, $(-1,\sqrt2)$, $(-2,\sqrt6)$, $(2,\sqrt2)$ and $(1,0)$, all centers on the real axis.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ia/paper-1-complex-circles.png]
{title=Five complex-plane circle loci, including the radius-zero point at alpha equals one}
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 satisfies
$$
|f(z)-c|=|f(z)-f(c)|=|z-c|,\qquad f(f(z))=z.
$$
Thus 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 \b[$f(C)=C$], including the surjectivity required by the question.
For any three nonempty <circles>, fix $u\in C_3$. The <triangle inequality> gives, for every $z\in C_1,w\in C_2$,
$$
|z-w|\le|z-u|+|u-w|\le m(C_1,C_3)+m(C_2,C_3).
$$
Taking the maximum over $z,w$ proves \b[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 $z=c_1+r_1v$ and $w=c_2-r_2v$, 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 $r_1+r_2$. Therefore
$$
\boxed{m(C_{\alpha_1\beta_1\gamma_1},C_{\alpha_2\beta_2\gamma_2})
=\frac{|\alpha_1\beta_2-\alpha_2\beta_1|}{|\alpha_1\alpha_2|}
+\frac{\sqrt{|\beta_1|^2-\alpha_1\gamma_1}}{|\alpha_1|}
+\frac{\sqrt{|\beta_2|^2-\alpha_2\gamma_2}}{|\alpha_2|}.}
$$
The function $m$ is not a <metric> on positive-radius <circles>, since $m(C,C)=2r$, even though it satisfies this triangle-type inequality.
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