= Solution
Differentiate $\mu\dot{\mathbf r}=\mathbf p+e^*\mathbf B\times\mathbf r$ and substitute the previous <momentum> equation:
$$
\mu\ddot{\mathbf r}=2e^*\mathbf B\times\dot{\mathbf r}
-\left(\frac{e^2B^2}{M}+\frac{e^2}{r^3}\right)\mathbf r.
$$
For a planar <vector>, $\mathbf B\times\dot{\mathbf r}=B(-\dot y,\dot x)$, which gives the two component equations with the printed signs.
For <circular relative motion of a neutral Coulomb pair> of separation radius $R$, $\dot{\mathbf r}=\omega\hat{\mathbf z}\times\mathbf r$ and $\ddot{\mathbf r}=-\omega^2\mathbf r$. Thus
$$
\mu\omega^2-2e^*B\omega-\frac{e^2B^2}{M}-\frac{e^2}{R^3}=0.
$$
Since $(e^*)^2+\mu e^2/M=e^2/4$, the two signed <frequencies> are
$$
\boxed{\omega_\pm=\frac{e^*B}{\mu}\pm
\sqrt{\frac{e^2B^2}{4\mu^2}+\frac{e^2}{\mu R^3}}.}
$$
One is positive and the other negative. At $B=0$ they give the two senses of an ordinary Coulomb circular orbit; equal <masses> remove the term linear in $B$ from their average. With negligible Coulomb attraction and $eB>0$, the branches approach $eB/m_2$ and $-eB/m_1$, the appropriate signed free cyclotron limits.
The actual particle paths follow by integrating the centre-of-mass <velocity>:
$$
\mathbf R_{\rm cm}=\mathbf R_0+\frac{eB}{M\omega}\mathbf r,
\quad
\mathbf r_1=\mathbf R_0+\frac{eB/\omega-m_2}{M}\mathbf r,
\quad
\mathbf r_2=\mathbf R_0+\frac{eB/\omega+m_1}{M}\mathbf r.
$$
Both particles move on circles about the same fixed <guiding centre> with a constant separation and common signed <frequency>, with a possible relative phase reversal encoded by a negative coefficient. Their mechanical <centre of mass> also generally circles. This is consistent with magnetic deflection plus a central mutual force; central interaction alone does not eliminate the external Lorentz forces.
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