= Solution
By <Newton's third law>, the <force> on the second particle is $-F_{12}$. For $M=m_1+m_2$ and <centre of mass> $R=(m_1r_1+m_2r_2)/M$, <Newton's second law> gives $M\ddot R=F_{12}-F_{12}=0$. Thus
$$
\boxed{R(t)=R(0)+t\dot R(0).}
$$
The centre moves on a straight line with constant <velocity>; zero <velocity> is allowed. For the relative position $r=r_1-r_2$, subtraction of the two equations gives
$$
\ddot r=\left(\frac1{m_1}+\frac1{m_2}\right)F_{12}(r),\qquad
\boxed{\mu\ddot r=F_{12}(r),\quad\mu=\frac{m_1m_2}{m_1+m_2}.}
$$
Hence the <two-body problem> reduces to a single particle of <reduced mass> $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.
\b[Yes: the two equal <masses> can follow the same fixed circle in diametrically opposite positions.] Put the stationary <centre of mass> at the circle's center and choose tangential <velocities> with the same sense of rotation. If the circle has radius $a$, the separation is $2a$ and the mutual gravitational <force> is directed toward that center, with magnitude $Gm^2/(4a^2)$. The <circular orbit> condition is $ma\omega^2=Gm^2/(4a^2)$, so an <equal-mass circular binary> has
$$
\boxed{\omega^2=\frac{Gm}{4a^3}.}
$$
These initial data maintain the opposite positions and give the same constant <angular speed> for both particles.
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