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By Newton's third law, the force on the second particle is $-F_{12}$. For $M=m_1+m_2$ and centre of mass $R=(m_1{r}_1+m_2{r}_2)/M$, Newton's second law gives $M\ddot{R}=F_{12}-F_{12}=0$. ThusThe 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 givesHence the two-body problem reduces to a single particle of reduced mass $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.

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 $G{m}^2/(4a^2)$. The circular orbit condition is $ma{\omega }^2=G{m}^2/(4a^2)$, so an equal-mass circular binary hasThese initial data maintain the opposite positions and give the same constant angular speed for both particles.