= Solution
Put $A(r)=r^2/(r-2m)^2$ and $B(r)=r^3/(r-2m)$, so $L=A\dot r^2+B\dot\varphi^2$. Direct differentiation gives $A'=-4mr/(r-2m)^3$ and $B'=2r^2(r-3m)/(r-2m)^2$. The <Euler-Lagrange equations> are $2A\ddot r+A'\dot r^2-B'\dot\varphi^2=0$ and $(d/dt)(2B\dot\varphi)=0$, equivalently
$$
\boxed{\ddot r-\frac{2m}{r(r-2m)}\dot r^2-(r-3m)\dot\varphi^2=0},
$$
$$
\boxed{\ddot\varphi+\frac{2(r-3m)}{r(r-2m)}\dot r\dot\varphi=0}.
$$
The cyclic angular coordinate and time <independence> give the two conserved quantities
$$
\boxed{J=\frac{2r^3}{r-2m}\dot\varphi,\qquad E=\frac{r^2}{(r-2m)^2}\dot r^2+\frac{r^3}{r-2m}\dot\varphi^2}.
$$
The 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 $(r_*-3m)\dot\varphi^2=0$, so $\boxed{r_*=3m}$. This lies outside $2m$, and the angular equation then gives $\varphi=\Omega t+\varphi_0$, 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.
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