= Solution
\b[The assertion about every exterior radius is false for geodesics.] A <circular orbit> here means an equatorial <null geodesic> at constant $r$, rather than a genuinely radial trajectory. A zero of the radial <effective potential> alone gives a turning point, not necessarily a <circular orbit>. The radial <geodesic equation> must hold as well.
Write $q=L-aE$ and
$$
f(r)=\dot r^2=E^2-\frac{q^2+2aEq}{r^2}+\frac{2Mq^2}{r^3}.
$$
For an exterior <circular null geodesic>, the two conditions are $f(r)=f'(r)=0$, equivalently $\mathcal R(r)=\mathcal R'(r)=0$ with $E,L$ fixed. One way to justify the derivative condition without dividing by $\dot r=0$ is to use the radial <Euler-Lagrange equation>. At constant $r$ it gives
$$
\partial_rg_{AB}\dot x^A\dot x^B=0,\qquad A,B\in\{t,\phi\}.
$$
Together with $g_{AB}\dot x^A\dot x^B=0$, and the fixed conserved $p_A$, this is precisely the double-root condition for $\mathcal R$.
Now
$$
f'(r)=\frac{2(q^2+2aEq)}{r^3}-\frac{6Mq^2}{r^4}.
$$
Setting this to zero and using $f=0$ gives
$$
q^2+2aEq=\frac{3Mq^2}{r},\qquad
E^2=\frac{Mq^2}{r^3}.
$$
A nontrivial exterior <null geodesic> cannot have $E=0$ here: the second equation would force $q=L=0$, and then every tangent component would vanish. Therefore
$$
\boxed{
r^2=\frac{L^2-a^2E^2}{3E^2},\qquad
r^3=\frac{M(L-aE)^2}{E^2}.
}
$$
Both equations, with the same constants $E,L$, are required. They give the requested radius relations in terms of the <conserved quantities>, while showing why an arbitrary exterior radius is not permitted.
For a more explicit classification let $b=L/E$ be the <impact parameter>. The second relation gives $b-a=s\,r^{3/2}/\sqrt M$ with $s=\pm1$. Substituting into the first relation yields
$$
\boxed{r^2-3Mr+2s\,a\sqrt{Mr}=0.}
$$
For $0<a<M$, $s=+1$ gives the prograde <equatorial circular Kerr photon orbit> and $s=-1$ the retrograde one. The two allowed exterior radii are
$$
\boxed{
\begin{aligned}
r_{\mathrm{pro}}
&=2M\left[1+\cos\left(\frac23\arccos(-a/M)\right)\right],\\
r_{\mathrm{retro}}
&=2M\left[1+\cos\left(\frac23\arccos(a/M)\right)\right].
\end{aligned}
}
$$
Both are strictly greater than $r_+$ in the subextremal range. Squaring the signed radius equation gives $r(r-3M)^2=4Ma^2$, but introduces an additional inner root and loses the sign distinguishing the two senses of rotation. At $a=0$ the two exterior radii coincide at the <photon sphere> $r=3M$, with $b=\pm3\sqrt3M$.
\Image[/past-exam-of-the-general-relativity-ii-course-of-the-university-of-oxford/2017-kerr-photon-orbit-radii.png]
{title=The two equatorial circular photon-orbit radii and the outer Kerr horizon as functions of spin}
For a direct counterexample to the all-radii claim take $a=0$ and $r_0=4M>r_+=2M$. Choose $E=1$ and $L=4\sqrt2M$ so that $f(r_0)=0$. The radial <geodesic equation> nevertheless gives
$$
\ddot r=\frac12f'(r)
=\frac{L^2(r-3M)}{r^4},
\qquad
\left.\ddot r\right|_{r=4M}=\frac1{8M}\ne0.
$$
It is a turning point, not a <circular orbit>. Alternatively, constant-$r$ null directions exist throughout the exterior because the two angular-velocity roots of $g_{tt}+2\Omega g_{t\phi}+\Omega^2g_{\phi\phi}=0$ are real; almost all of their integral curves require acceleration and fail the radial <geodesic equation>. This distinction explains how an all-radii assertion can arise.
Finally, $r=r_+$ itself has the <null geodesic> generators of its <Killing horizon>, proportional to $\chi_+=\partial_v+\Omega_+\partial_\psi$. They are a separate horizon case. On the horizon $\chi_+$ is normal to every tangent, so its contractions with $\partial_v$ and $\partial_\psi$ both vanish: $E=L=0$ for a generator tangent. Its <affine parameter> can be chosen by rescaling that tangent. The exterior calculation dividing by $E$ does not include it. The presence of these generators does not supply <circular null geodesics> at the intervening arbitrary exterior radii.
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