= Solution
For the static nonrotating source, use <spherical symmetry> and the <static scalar perturbation of Minkowski spacetime>, with isotropic radius $\rho$:
$$
ds^2=(1+2A)dt^2-(1+2C)(d\rho^2+\rho^2d\Omega^2).
$$
To first order, the <Einstein tensor> in the declared convention has
$$
\delta G_{00}=2\Delta C,\qquad\delta G_{ij}=\partial_i\partial_j(A+C)-\delta_{ij}\Delta(A+C).
$$
These are also the <static scalar perturbation of Minkowski spacetime> equations in the original information sheet. The source has $T_{00}=M\delta^{(3)}(\mathbf x)$ and zero spatial stress. With the sign established in Question 2,
$$
\Delta C=-4\pi GM\delta^{(3)}(\mathbf x).
$$
The <distributional identity> $\Delta(1/\rho)=-4\pi\delta^{(3)}(\mathbf x)$ and <asymptotic flatness> give $C=GM/\rho$. For $F=A+C$, the spatial field equation has trace $-2\Delta F=0$ and therefore $\partial_i\partial_jF=0$. Its decaying spherical solution is $F=0$, so $A=-GM/\rho$. This determines both potentials; the spatial correction cannot be inferred from the Newtonian <gravitational potential> alone.
Now use the <isotropic-to-areal weak-field radius change>. The angular coefficient defines $r^2=(1+2GM/\rho)\rho^2$, hence $r=\rho+GM$ to first order. Thus $dr=d\rho$ to the same accuracy, and replacing $1/\rho$ by $1/r$ inside a perturbation changes only second-order terms. With the <Schwarzschild radius> $r_S=2GM$, the <linearized Schwarzschild metric> is
$$
\boxed{ds^2=(1-r_S/r)dt^2-(1+r_S/r)dr^2-r^2(d\theta^2+\sin^2\theta\,d\phi^2).}
$$
This is an exterior <weak-field approximation>: $r_S/r\ll1$. The point source is used distributionally to fix the mass, not to claim a regular or weak field at the origin.
For a <null geodesic> choose its orbital plane as $\theta=\pi/2$, using rotational symmetry, and let a dot denote an <affine parameter> derivative. For a <Killing vector> $k$ and an affinely parametrized <geodesic> tangent $U$, the <geodesic conserved quantity from a Killing vector> follows directly from
$$
\frac{d}{d\lambda}(k_aU^a)=U^aU^b\nabla_{(a}k_{b)}+k_a\nabla_UU^a=0.
$$
Applying this to the time-translation and axial <Killing vectors>, with the spatial sign absorbed into the definition of $h$, gives
$$
\boxed{E=(1-u)\dot t,\qquad h=r^2\dot\phi,\qquad u=r_S/r.}
$$
If the tangent is normalized as the photon's <four-momentum>, $E$ is its energy measured at infinity and $h$ its signed <angular momentum> about the selected axis. Rescaling an arbitrary <affine parameter> rescales both; the ratio $b=|h|/E$ is the invariant <impact parameter>. Future-directed orbits have $E>0$. Assume $h\ne0$ to use $\phi$ as the orbit parameter; $h=0$ gives radial rays, not a finite-radius circular orbit.
The <null vector> constraint gives
$$
0=\frac{E^2}{1-u}-(1+u)\dot r^2-\frac{h^2}{r^2},\qquad \dot r^2=\frac{E^2}{1-u^2}-\frac{h^2}{r^2(1+u)}.
$$
Expand consistently to first order in the <weak-field approximation> correction: $(1-u^2)^{-1}=1+O(u^2)$ and $(1+u)^{-1}=1-u+O(u^2)$. Also $du/d\phi=-r_S\dot r/h$. Therefore the first-order orbital equation is
$$
\boxed{\left(\frac{du}{d\phi}\right)^2=\frac{r_S^2E^2}{h^2}-u^2(1-u).}
$$
The cubic term is the first correction relative to the flat inverse-radius term; the equation is not a controlled exact equation for large $u$.
A constant solution of a <first integral> is not by itself sufficient to establish a circular <geodesic>. For a circular trajectory, the radial <Euler-Lagrange equation> also requires $A'(r)\dot t^2/2=r\dot\phi^2$, where now $A(r)=1-r_S/r$ is the metric's time coefficient. Combining this with the <null vector> constraint gives $rA'=2A$, or $u=2/3$. Equivalently, the formal potential $V(u)=u^2(1-u)$ must have a stationary point. Apart from the infinite-radius endpoint $u=0$, its only stationary point is $u_c=2/3$, with $V(u_c)=4/27$. Thus the formal circular solution requires
$$
\boxed{r_c=\frac32r_S,\qquad\left|\frac Eh\right|=\frac{2}{3\sqrt3\,r_S},\qquad b_c=\frac{3\sqrt3}{2}r_S.}
$$
The reduced radial equation is $u''=-u+\tfrac32u^2$, with prime denoting $d/d\phi$. It follows by differentiating the <first integral> on moving segments and agrees with the regular <Euler-Lagrange equation> at turning points. Perturbing $u=u_c+\epsilon$ gives
$$
\epsilon''=\epsilon+O(\epsilon^2),\qquad\epsilon=A_+e^\phi+A_-e^{-\phi}+O(\epsilon^2).
$$
There is an exponentially growing <linear instability> mode; equivalently $V''(u_c)=-2<0$. \b[The formal circular orbit is unstable.]
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-60-photon-potential.png]
{title=The formal inverse-radius photon potential has its maximum outside the weak-field regime}
The <weak-field photon-orbit validity> limitation is essential: $u_c=2/3$ is not small. There is therefore \b[no finite-radius circular photon orbit established within the controlled weak-field region]. The formal extrapolation happens to reproduce the radius and <impact parameter> of the exact Schwarzschild <photon sphere>, but that physical result needs the exact geometry. The calculation also cannot establish a <Schwarzschild event horizon> from a first-order expansion at $u\sim1$.
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