= Solution
Take $M>0$, set $b=Q^2/M$, and denote the squared areal radius by $C(r)=r(r-b)$. The positive-area exterior has $r>b$. The stationary norm is $-f(r)$, where $f(r)=1-2M/r$, so its candidate <Killing horizon> is at $r=2M$. There it has area
$$
A_H=4\pi C(2M)=8\pi(2M^2-Q^2).
$$
A regular horizon must have $b<2M$. The <curvature singularity> then lies strictly inside the horizon, and ingoing coordinates extend smoothly across it because $C(2M)>0$. In the future interior $f<0$, the gradient of $r$ is timelike and future causal curves move to decreasing $r$, rather than back to infinity. Thus the regular <black hole> condition is
$$
\boxed{M>0,\qquad Q^2<2M^2.}
$$
The equality case has zero horizon area and a curvature-singular null limiting surface. It is often included in the family as a singular extremal solution, but it is not a regular <black hole>. If that limiting nomenclature is used, the family condition is $Q^2\leq2M^2$ with this essential qualification. This geometry has the same spherical form as the <magnetically charged dilaton black hole>; no unprinted field equation is needed for the horizon comparison.
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