= Solution
First supply the introductory scaling and geometry. The leading tangential <Stokes equation> in a thin gap balances $P/L$ against $\mu U/H^2$. The <shear stress> scale is $\tau\sim\mu U/H$, so <lubrication pressure dominates shear stress>:
$$
\boxed{P\sim\frac{\mu UL}{H^2},\qquad \frac\tau P\sim\frac HL\ll1.}
$$
The outer circle, viewed along a ray $r(\cos\theta,\sin\theta)$ from the inner centre, satisfies $r^2-2\alpha\Delta r\sin\theta+\alpha^2\Delta^2=(a+\Delta)^2$. Expanding its positive root to first order in $\Delta/a$ gives $r=a+\Delta+\alpha\Delta\sin\theta+O(\Delta^2/a)$, hence $h=\Delta(1+\alpha\sin\theta)$.
Take positive $\Omega$ to mean increasing $\theta$, and put $s=a\theta$, $y=r-a$, $U=\Omega a$. With <no-slip boundary conditions> at $y=0,h$, <lubrication theory> gives
$$
u_\theta=U\left(1-\frac yh\right)+\frac{p_s}{2\mu}y(y-h),\qquad
q=\frac{Uh}{2}-\frac{h^3}{12\mu}p_s.
$$
Steady <mass conservation> makes $q$ independent of $\theta$. Single-valued <pressure> requires $\int_0^{2\pi}p_\theta\,d\theta=0$. Write $I_n=\int_0^{2\pi}(1+\alpha\sin\theta)^{-n}d\theta$ and $D_\alpha=1+\alpha^2/2$. It follows that
$$
\boxed{q=\frac{\Omega a\Delta}{2}\frac{I_2}{I_3}
=\frac{\Omega a\Delta}{2}\frac{1-\alpha^2}{D_\alpha},\qquad
p_s=\frac{6\mu\Omega a}{h^2}-\frac{12\mu q}{h^3}.}
$$
The tangential fluid tractions on the inner and outer solids, measured in the same increasing-$\theta$ direction, are respectively
$$
\boxed{\tau_i=\mu u_{\theta,y}(0)=\frac{6\mu q}{h^2}-\frac{4\mu\Omega a}{h},\qquad
\tau_o=-\mu u_{\theta,y}(h)=\frac{6\mu q}{h^2}-\frac{2\mu\Omega a}{h}.}
$$
This sign convention distinguishes the traction on a wall from the corresponding stress component in the fluid.
To leading order the holding <force> balances the <pressure> traction on the inner circle. <Integration by parts> gives
$$
F_x=a\int_0^{2\pi}p\cos\theta\,d\theta
=-a\int_0^{2\pi}p_\theta\sin\theta\,d\theta.
$$
Use $p_\theta=a p_s$ and $\int\sin\theta/(1+\alpha\sin\theta)^n\,d\theta=(I_{n-1}-I_n)/\alpha$, with the continuous limit at $\alpha=0$. Substituting the boxed flux yields
$$
\boxed{F_x=-\frac{6\pi\mu\Omega a^3}{\Delta^2}\frac{\alpha}{\sqrt{1-\alpha^2}\,D_\alpha},\qquad F_z=0.}
$$
The zero vertical component also follows from <reflection in mathematics> in the vertical line: this leaves the geometry and vertical <force> unchanged, reverses $\Omega$, and <Linearity of Stokes flow> then requires $F_z=0$.
<Pressure> gives no moment about either circular wall's own centre. Integrating the <viscous shear torque> with lever arm $a$ to leading order gives the fluid couples
$$
\boxed{C_i=-\frac{2\pi\mu\Omega a^3}{\Delta\sqrt{1-\alpha^2}}\frac{1+2\alpha^2}{D_\alpha},\qquad
C_o=\frac{2\pi\mu\Omega a^3\sqrt{1-\alpha^2}}{\Delta D_\alpha}.}
$$
Indeed $C_i=a^2\int\tau_i d\theta$ and $C_o=a^2\int\tau_o d\theta$ at this order. Applied holding or driving couples are the negatives of these fluid couples. They need not cancel because they are moments about different axes. The fluid <force> on the outer solid equals the holding <force> $F_x$ on the inner one. Transferring its moment to the inner axis adds $-\alpha\Delta F_x$, so <angular momentum> balance about a common axis is
$$
\boxed{C_i+C_o=\alpha\Delta F_x.}
$$
For $\alpha=0$ the usual equal and opposite concentric couples are recovered. All these formulas are leading thin-gap expressions at fixed $|\alpha|<1$; taking the contact limit requires maintaining the local slender-gap assumptions.
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