= Solution
The <Milne-Thomson circle theorem>[circle theorem] says that if a <complex potential of a two-dimensional potential flow>[complex potential] $f$ has no singularities on or inside $|z|=a$ (and has the required analytic continuation there), then
$$
w(z)=f(z)+\overline{f(a^2/\bar z)}
$$
adds image singularities inside the circle and makes it a <streamline>. On the circle $a^2/\bar z=z$, so $w=f(z)+\overline{f(z)}$ is real and its imaginary part is constant. This gives zero normal velocity. The reflected term is holomorphic where defined, as follows from its <power function of a statistical test>[power] series in $1/z$, and its singularities are precisely the reflected ones; the external original singularities are retained. This proves the construction, with an additive constant and any independently prescribed <circulation> handled separately.
For the <fluid vortex> use $f(z)=-i\Gamma\log(z-b)/(2\pi)$. Reflecting gives an opposite image at $d=a^2/b$ and also a term $-i\Gamma\log z/(2\pi)$, up to constants. The zero-<circulation> circle-theorem construction has both these terms. To impose the specified <circulation> $-\Gamma$ about the obstacle, add $+i\Gamma\log z/(2\pi)$, canceling the central term. Thus
$$
\boxed{w(z)=\frac{i\Gamma}{2\pi}\bigl(\log(z-d)-\log(z-b)\bigr)+\text{constant}.}
$$
It has the stated <circulation> and the same impermeable boundary, since the added central-<fluid vortex> <streamfunction> is constant on the circle.
For the given <Blasius force theorem>[Blasius integral] take the obstacle contour counterclockwise. Inside it the sole singularity of
$$
w'(z)^2=-\frac{\Gamma^2}{4\pi^2}\left(\frac1{z-d}-\frac1{z-b}\right)^2
$$
is at $d$. Its <residue> is $\Gamma^2/[2\pi^2(d-b)]$, from the cross term. The <residue theorem> then gives
$$
F_x-iF_y=-\rho\pi\operatorname{Res}_{z=d}w'^2=\frac{\rho\Gamma^2}{2\pi(b-d)}.
$$
Therefore $\boxed{F_x=\rho\Gamma^2b/[2\pi(b^2-a^2)],\quad F_y=0}$. The force on the obstacle points toward the external <fluid vortex>.
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