Solution
= Solution
For the proposed <circular motion>,
$$
\ddot{\mathbf x}_1=-a\omega^2(\cos\omega t,\sin\omega t,0).
$$
The separation is $\mathbf x_1-\mathbf x_2=2a(\cos\omega t,\sin\omega t,0)$, so the <inverse-square force> on particle 1 is
$$
\mathbf F_{12}
=-\frac{km^2}{4a^2}(\cos\omega t,\sin\omega t,0).
$$
The equation $m\ddot{\mathbf x}_1=\mathbf F_{12}$ holds exactly when
$$
\boxed{\omega^2=\frac{km}{4a^3}}.
$$
The equation for particle 2 follows by symmetry.