I will assume the phrase "circular orbits around each other" means that you are looking for a solution where the two bodies remain constant distance apart.
The easiest thing to do is to work in a coordinate system adapted to the centre of mass. The total separation between the two masses is $R = r_1 + r_2$. The centre of mass sits at $(R_0,0)$ where $R_0 = (-m_1 r_1 + m2 r_2)/(m_1 + m_2)$. Without loss of generality, we'll assume that the centre of mass remains immobile (else we can subtract the velocity of the centre of mass from both bodies and use the fact that Galilean motions leave the mechanics invariant).
So the distance from $m_1$ to the centre of mass is $|-r_1 - R_0| = R m_2 / (m_1 + m_2)$, similarly the distance from $m_2$ to the centre of mass will be $R m_1/(m_1 + m_2)$.
Now, let us write down the equations of motion.
The assumption that the centre of mass remains immobile implies that the total momentum is 0. This means that $m_1 v_1 + m_2 v_2 = 0$. Furthermore, that $R$ is constant requires that the initial velocities be such that the two masses start out moving perpendicular to the $x$ axis. So hereon we can assume that $v_1, v_2$ are scalars, which represent the initial speed in the $y$ direction for the masses respectively.
Relative to the (immobile) centre of mass, the first mass is travelling in a circular orbit of radius $R m_2 / (m_1 + m_2)$. This requires force $ m_1 v_1^2 / (R m_2 / (m_1 + m_2) ) = v_1^2 \cdot \frac{m_1 (m_1 + m_2)}{R m_2}$. We have also a similar expression for the force acting on $m_2$.
The gravitational attraction between the two masses is $G m_1 m_2 / R^2$.
So now we solve the system
$$ m_1 v_1 + m_2 v_2 = 0 $$
$$ \frac{G m_1 m_2}{R^2} = v_1^2 \frac{m_1 (m_1 + m_2)}{m_2 R} = v_2^2 \frac{m_2(m_1 + m_2)}{m_1R} $$
Which gives
$$ v_1 = \sqrt{ \frac{G m_2^2}{R(m_1 + m_2)} }, v_2 = \sqrt{ \frac{G m_1^2}{R(m_1 + m_2)} }$$