 Solution

Put $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$, so $L=A{\dot{r}}^2+B{\dot{\phi }}^2$. Direct differentiation gives $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The Euler-Lagrange equations are $2A\ddot{r}+A^{\prime }{\dot{r}}^2-B^{\prime }{\dot{\phi }}^2=0$ and $(d/dt)(2B\dot{\phi })=0$, equivalentlyThe cyclic angular coordinate and time independence give the two conserved quantitiesThe expression for $E$ equals $L$ here because $L$ is homogeneous quadratic in the velocities.

For a constant radius and nonzero angular motion, the radial equation requires $(r_{\ast }-3m){\dot{\phi }}^2=0$, so $\boxed{r_{\ast }=3m}$. This lies outside $2m$, and the angular equation then gives $\phi =\Omega t+{\phi }_0$, with any nonzero constant $\Omega$. This is the photon sphere radius in the stated coordinates. If both velocities are zero, the displayed variational equations also admit a stationary configuration at any radius; the moving circular photon solution is the nonzero-speed case just found.