= Solution
Put $r_+=\ell\sqrt M$ and $f=(r^2-r_+^2)/\ell^2$. The apparent failure of the nonrotating <BTZ black hole> coordinates at $r=r_+$ is a <coordinate singularity>. Define the <tortoise coordinate> and ingoing time by
$$
r_* =\frac{\ell^2}{2r_+}\log\left|\frac{r-r_+}{r+r_+}\right|,
\qquad v=t+r_*.
$$
The <Ingoing BTZ coordinates> give
$$
ds^2=-f\,dv^2+2\,dv\,dr+r^2d\phi^2.
$$
All coefficients are analytic at $r_+>0$, and the determinant is $-r^2$, so the <metric tensor> extends nondegenerately through that radius. The <Killing vector field> $K=\partial_v$ agrees with $\partial_t$ in the exterior. On $r=r_+$, $K^2=-f=0$ and $K_a=(dr)_a$. It is therefore both normal and tangent to this <null hypersurface>, proving that it is a <Killing horizon>.
One may include both horizon branches and their <bifurcation surface> explicitly. Set $\kappa=r_+/\ell^2$, $u=t-r_*$ and, initially in the exterior, $U=-e^{-\kappa u}$, $V=e^{\kappa v}$. Then
$$
UV=-\frac{r-r_+}{r+r_+},\qquad
r=r_+\frac{1-UV}{1+UV},
\qquad
ds^2=-\frac{4\ell^2\,dU\,dV}{(1+UV)^2}+r_+^2\left(\frac{1-UV}{1+UV}\right)^2d\phi^2.
$$
This <analytic extension of a spacetime> is regular near $UV=0$, and $K=\kappa(V\partial_V-U\partial_U)$ extends analytically. The surfaces $U=0$ and $V=0$ form the \b[extended Killing horizon at $r=r_+$]; $K$ vanishes only at their regular bifurcation circle.
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