= Solution
The separation of the two bodies is $a$, as is fixed by the supplied <Kepler third law> $\omega_o^2=G(M_1+M_2)/a^3$. The phrase “orbit of radius $a$” cannot denote the first body's distance from the center of mass in that formula.
Expand the companion's <Newtonian potential of a point mass> about the first body's center. The constant term produces no force; the linear term produces the uniform acceleration of the center and disappears in the translating frame. The leading differential acceleration therefore comes from the <quadrupolar tidal forcing>
$$
\Phi_{\mathrm{tide}}=-\frac{GM_2}{2a^3}\left[3(\mathbf n(t)\cdot\mathbf r)^2-r^2\right],
$$
where $\mathbf n(t)$ points toward the companion. It is a degree-two solid <spherical harmonic>. For a circular nonrotating-frame orbit, its time-dependent components have $m=\pm2$ and frequency $2\omega_o$, because the squared direction cosines repeat twice per orbit. There is also a static $m=0$ component. The real part of one complex harmonic represents the pair of time-dependent components; its normalization and orientation are absorbed into the order-one coefficient $\psi$. This gives $\Psi=\psi(GM_2/a^3)r^2Y_2^m$ and $\omega=2\omega_o$. The expansion is the leading term for $R_1/a$ small; spin is being neglected.
For a forced <potential flow> with the same temporal frequency, the <Euler momentum equation> becomes
$$
-\omega^2\nabla U=-\nabla\left(\frac{\delta p}{\rho_1}+\delta\Phi+\Psi\right).
$$
Use $U=A r^2Y_2^m$. The <surface gravity perturbation of a uniform-density sphere> is unchanged, as is the vanishing <Lagrangian pressure perturbation> boundary condition. Integrating the equation and applying that condition gives
$$
(\omega^2-\omega_2^2)A=\psi\frac{GM_2}{a^3},\qquad
\omega_2^2=\frac45\frac{GM_1}{R_1^3}.
$$
Since $\xi_r(R_1)=2AR_1Y_2^m$, the real surface <displacement> is
$$
\boxed{\xi_r(R_1,t)=\epsilon R_1\frac{\omega_2^2}{\omega^2-\omega_2^2}
\operatorname{Re}\!\left(\psi Y_2^m e^{-i\omega t}\right),\qquad
\epsilon=\frac52\frac{M_2}{M_1}\left(\frac{R_1}{a}\right)^3.}
$$
Indeed $\epsilon\omega_2^2=2GM_2/a^3$, so the normalization is consistent. This is a particular forced response away from the <resonance>; arbitrary free <normal modes> can be added.
Put $\mu=M_2/M_1$ and $q=(R_1/a)^3$. The <tidal resonance of a stellar oscillation> condition is $4\omega_o^2=\omega_2^2$, hence
$$
q=\frac1{5(1+\mu)},\qquad
\epsilon=\frac{\mu}{2(1+\mu)},\qquad
\boxed{\left(\frac{R_1}{a}\right)^3=\frac15(1-2\epsilon).}
$$
The resonance gives $0\leq\epsilon<1/2$ for finite nonnegative $\mu$, with actual forcing requiring $M_2>0$. At exact <resonance> the undamped <linearization> has no bounded response at the forcing frequency: the time-domain particular solution grows proportionally to $t$ times a sinusoid. Thus the frequency-response formula has a pole, rather than describing a finite resonant <displacement>. Away from that pole, <linearization> requires the displayed surface displacement to be small compared with $R_1$; the leading <quadrupolar tidal forcing> approximation separately requires $R_1/a$ small.
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