= Solution
Set $\mathbf r=\mathbf x_1-\mathbf x_2$, $M=M_1+M_2$ and $\mu=M_1M_2/M$, the <reduced mass>. <Newton's law of universal gravitation> gives $\ddot{\mathbf x}_1=-GM_2\mathbf r/r^3$ and $\ddot{\mathbf x}_2=GM_1\mathbf r/r^3$. Subtraction establishes
$$
\boxed{\ddot{\mathbf r}=-\frac{GM}{r^3}\mathbf r.}
$$
For fixed masses, the orbital <energy> and <angular momentum> are
$$
E=\frac{\mu}{2}\mathbf v^2-\frac{GM\mu}{r},\qquad \mathbf J=\mu\mathbf r\times\mathbf v,\qquad \mathbf v=\dot{\mathbf r}.
$$
Taking derivatives gives $\dot E=\mu\mathbf v\cdot(\ddot{\mathbf r}+GM\mathbf r/r^3)=0$ and $\dot{\mathbf J}=\mu\mathbf r\times\ddot{\mathbf r}=0$. These are <conservation of energy> and <conservation of angular momentum> in the unperturbed <Kepler orbit>. To see directly that the <orbital eccentricity> is fixed, put $\mathbf h=\mathbf r\times\mathbf v$ and $\mathbf n=\mathbf r/r$. The <eccentricity vector> is
$$
\mathbf e=\frac{\mathbf v\times\mathbf h}{GM}-\mathbf n.
$$
Since $\dot{\mathbf h}=0$, the <cross product> identity $\mathbf r\times(\mathbf r\times\mathbf v)=\mathbf r(\mathbf r\cdot\mathbf v)-r^2\mathbf v$ gives $\ddot{\mathbf r}\times\mathbf h=GM\dot{\mathbf n}$, so $\dot{\mathbf e}=0$. Its magnitude $e$ is the <orbital eccentricity>. Equivalently,
$$
\boxed{e^2=1+\frac{2EJ^2}{G^2M^2\mu^3},\qquad E=-\frac{GM\mu}{2a}.}
$$
Thus the <semi-major axis> and <orbital eccentricity> of a bound noncollision <Kepler orbit> are constant, including $e=0$.
Take the <centre of mass> as origin. The two positions are $\mathbf x_1=(M_2/M)\mathbf r$ and $\mathbf x_2=-(M_1/M)\mathbf r$. For $x\gg r$, a <Taylor expansion> gives
$$
\frac1{|\mathbf x-\mathbf b|}=\frac1x+\frac{\mathbf x\cdot\mathbf b}{x^3}+\frac{3(\mathbf x\cdot\mathbf b)^2-x^2b^2}{2x^5}+O\left(\frac{b^3}{x^4}\right).
$$
The mass dipole vanishes because $\sum_A M_A\mathbf x_A=0$, and the <second mass moment tensor> is $\sum_A M_Ax_{Ai}x_{Aj}=\mu r_i r_j$. The <gravitational quadrupole potential of a point-mass binary> is consequently
$$
\Phi(\mathbf x)=-\frac{GM}{x}-\frac{G\mu}{2x^5}\bigl[3(\mathbf x\cdot\mathbf r)^2-x^2r^2\bigr]+O\left(\frac{GMr^3}{x^4}\right).
$$
Both printed tensors are trace-free. Contracting them with the <Kronecker delta> summed over three spatial dimensions gives
$$
q_{ij}l_{ij}=\frac{\mu}{2x^5}\bigl[3(\mathbf x\cdot\mathbf r)^2-x^2r^2\bigr].
$$
This proves the requested expression through second order. In particular the printed convention is $q_{ij}=\tfrac32 Q_{ij}$, where $Q_{ij}=\mu(r_ir_j-r^2\delta_{ij}/3)$ is the standard <mass quadrupole moment>; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in <centre of mass> coordinates.
To evaluate the <quadrupole formula> without assuming a <circular orbit>, put $k=GM$, $u=\dot r=\mathbf n\cdot\mathbf v$ and $\mathbf v_\perp=\mathbf v-u\mathbf n$. Differentiating the <gravitational acceleration> gives the <kinematic jerk>
$$
\dddot{\mathbf r}=-\frac{k}{r^3}(\mathbf v-3u\mathbf n).
$$
For $S_{ij}=r_ir_j$, the product rule then gives
$$
\dddot S_{ij}=\frac{k}{r^2}\bigl[-4(n_iv_j+v_in_j)+6u n_i n_j\bigr],\qquad \frac{d^3r^2}{dt^3}=-\frac{2ku}{r^2}.
$$
Inserting these into the printed <mass quadrupole moment> convention yields
$$
\dddot q_{ij}=\frac{\mu k}{r^2}\bigl[u(\delta_{ij}-3n_in_j)-6(n_iv_{\perp j}+v_{\perp i}n_j)\bigr].
$$
The two tensors in brackets are orthogonal in their <tensor contraction>, because $\mathbf n\cdot\mathbf v_\perp=0$. Their squared norms are $6$ and $2v_\perp^2$, respectively. Hence
$$
\dddot q_{ij}\dddot q_{ij}=\frac{\mu^2k^2}{r^4}(6u^2+72v_\perp^2)=\frac{72\mu^2k^2}{r^4}\left(\mathbf v^2-\frac{11}{12}u^2\right).
$$
The <instantaneous quadrupole luminosity of a Kepler binary> is therefore
$$
\boxed{\dot E=-\frac{32G^3M^2\mu^2}{5c^5r^4}\left(\dot{\mathbf r}\cdot\dot{\mathbf r}-\frac{11}{12}\dot r^2\right).}
$$
The radial derivative $\dot r$ is not the vector speed. The bracket equals $v_\perp^2+u^2/12$, so the radiated <luminosity> is nonnegative for every instantaneous velocity.
For a slowly evolving <circular orbit>, there is no preferred orbital phase at which to excite a persistent <eccentricity vector>. More precisely, the rotating <mass quadrupole moment> emits at twice the orbital frequency, with angular harmonic number two; its <gravitational-wave energy and angular-momentum balance> is $\dot E=\Omega\dot J$. The circular <Kepler orbit> sequence has $J=\mu\sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular <orbital eccentricity>. This is a <circular gravitational-wave inspiral>, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The <energy> flux alone would not establish circularity without this symmetry and <angular momentum> balance.
To leading radiation order use $r=a$, $u=0$ and $\mathbf v^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu/(2a)$ gives
$$
\frac{GM\mu}{2a^2}\dot a=-\frac{32G^4M^3\mu^2}{5c^5a^5},\qquad
\boxed{\frac{\dot a}{a}=-\frac{64G^3M^2\mu}{5c^5a^4}.}
$$
Keeping the fixed masses and leading <quadrupole formula>, integration gives
$$
a(t)^4=a_0^4-\frac{256G^3M^2\mu}{5c^5}(t-t_0),\qquad
\boxed{t_{\rm coal}-t_0=\frac{5c^5a_0^4}{256G^3M^2\mu}.}
$$
This formal coalescence time displays the strong $a_0^4$ dependence: <gravitational-wave emission from a binary system> matters far more for close <binary stars> than for wide ones. The <orbital period> decreases as $a^{3/2}$ and the <gravitational-wave frequency> increases, producing a chirp. A <detached binary> can be driven into <Roche-lobe overflow>; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.
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