= Solution
Since the binary separation and <mean motion> are both one, the <Kepler third law> gives
$$
\mu_1+\mu_2=G(M_1+M_2)=1.
$$
The <center of mass> condition puts the bodies at
$$
\mathbf r_{M_1}=(-\mu_2,0),
\qquad
\mathbf r_{M_2}=(\mu_1,0)
$$
in the <rotating reference frame>. Thus the sketch has $M_1$, $O$, and $M_2$ in that order along the $x$ axis, with $P=(x,y)$, and
$$
\boxed{OM_1=\mu_2,\qquad OM_2=\mu_1},
$$
$$
\boxed{M_1P=r_1=\sqrt{(x+\mu_2)^2+y^2}},
$$
$$
\boxed{M_2P=r_2=\sqrt{(x-\mu_1)^2+y^2}}.
$$
In the inertial sketch this entire configuration rotates uniformly about $O$.
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