 Solution

For $h\ne 0$, the effective potential vanishes at $r=2\mu$, rises to a maximum, and then falls to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=h^2(3\mu -r)/r^4$, so the only circular null geodesic hasThe second derivative there is $V_{\mathrm{eff}}^{\prime \prime }=-h^2/(81{\mu }^4)<0$, 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 $r\downarrow 0$. For purely radial light, $h=0$, the potential is identically zero and there is no circular orbit.