= Solution
For $h\ne0$, the <effective potential> vanishes at $r=2\mu$, rises to a maximum, and then falls to zero as $r\to\infty$. Its <derivative> is $V'_{\rm eff}=h^2(3\mu-r)/r^4$, so the only circular <null geodesic> has
$$
\boxed{r=3\mu,\qquad V_{\rm eff}=\frac{h^2}{54\mu^2}.}
$$
The second derivative there is $V''_{\rm eff}=-h^2/(81\mu^4)<0$, proving instability by the <effective potential stability criterion>. These circular orbits make up the <photon sphere>. If the formula is continued inside the horizon, the potential is negative and diverges to $-\infty$ as $r\downarrow0$. For purely radial light, $h=0$, the potential is identically zero and there is no circular orbit.
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2022-photon-potential-and-escape-cone.png]
{title=Schwarzschild null radial potential and the static-observer escape cone inside the photon sphere}
{description=Original dimensionless plots. The left panel shows the exterior potential and its maximum at $r=3\mu$; the right panel shows the critical escape half-angle for $2\mu<r\le3\mu$. The horizon endpoint is a limit of exterior static observations.}
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