= Solution
Now the eccentric <planet> $c$ is inside the circular perturber $d$. The <interior eccentricity-type mean-motion resonance> has
$$
\boxed{\phi_{cd}=(m+n)\lambda_d-m\lambda_c-n\varpi_c}.
$$
In the mean-conjunction approximation, $\phi_{cd}=n(\Lambda-\varpi_c)$, so $\Lambda_j=\varpi_c+(\phi_{cd}+2\pi j)/n$. Successive <conjunctions> advance by $2\pi m/n$; coprime $m,n$ visit $n$ distinct phases.
The near-conjunction tangential pulls are reversed for an eccentric body inside its perturber. Just before <conjunction> the inner body is pulled forwards and afterwards backwards. Near <periapsis>, an inward-moving inner body experiences the stronger earlier forward pull; its <semi-major axis> grows, its <mean motion> falls, and $\phi_{cd}$ increases because its coefficient of $\lambda_c$ is negative. On the outward side the feedback reverses, restoring the first-order centre zero. Higher-order feedback is strongest on the side closest to the external perturber, near <apoapsis>; the leading coefficient alternates in sign with order.
Writing $R_{\rm res}=B_ne_c^n\cos\phi_{cd}$ gives $B_n<0$ for odd $n$ and $B_n>0$ for even $n$ in the leading non-crossing model. The same negative-curvature <Hamiltonian> argument therefore gives
$$
\boxed{\phi_{cd,0}=\begin{cases}0,&n=1,3,\\ \pi,&n=2.\end{cases}}
$$
Conjunction phases relative to <periapsis> are $\{0\}$, $\{\pi/2,3\pi/2\}$ and $\{0,2\pi/3,4\pi/3\}$, respectively, as shown in the lower row of the figure. As before, these are small-eccentricity symmetric predictions; additional resonant <Fourier harmonics> and coupled secular dynamics can change the stable branches.
For actual alignment of the eccentric body at true longitude $\Lambda$, the exact phase relation is instead $\phi_{cd}=(m+n)x-my$, where $x=\Lambda-\varpi_c$ and $y=M_c(x)$. The simple equally spaced angles replace $y$ by $x$ at leading order.
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