= Solution
{c}
If $L$ is independent of $q_1$, the <Euler-Lagrange equation> makes the generalized momentum $p_1=\partial L/\partial\dot q_1$ constant. If $L$ is independent of time, the <energy function> $E=\sum_a\dot q_a(\partial L/\partial\dot q_a)-L$ is constant; differentiating it gives $\dot E=-L_t$ along solutions.
Set $A(r)=r^2/(r-2m)^2$ and $B(r)=r^3/(r-2m)$. Their derivatives are $A'=-4mr/(r-2m)^3$ and $B'=2r^2(r-3m)/(r-2m)^2$. The radial and angular equations are
$$
2A\ddot r+A'\dot r^2-B'\dot\phi^2=0,\qquad
\frac d{dt}(2B\dot\phi)=0.
$$
Equivalently,
$$
\ddot r-\frac{2m}{r(r-2m)}\dot r^2-(r-3m)\dot\phi^2=0.
$$
Two conserved quantities are
$$
\boxed{\ell=2B\dot\phi,\qquad E=A\dot r^2+B\dot\phi^2.}
$$
The second equals $L$ here because the Lagrangian is homogeneous quadratic in velocities. For a constant-radius orbit with nonzero angular motion, the radial equation requires $r-3m=0$. Thus
$$
\boxed{r_*=3m.}
$$
It lies outside $2m$ and is the circular photon orbit. If both velocities were zero, the differential equations would also admit static constant configurations at any radius; these are degenerate zero-energy paths rather than the intended moving photon trajectory.
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