= Solution
{c}
<Milne-Thomson's circle theorem> states that if $f(z)$ is a <complex potential of a two-dimensional potential flow> whose singularities lie outside $|z|=a$, then
$$
w(z)=f(z)+\overline{f(a^2/\bar z)}
$$
has the circle as a <streamline>, with the original exterior singularities unchanged. Additive constants in the potential are immaterial.
Take $f(z)=\Gamma(2\pi i)^{-1}\log(z-z_1)$, using the convention that positive $\Gamma$ is positive <circulation>. Reflection in the circle gives, up to a constant,
$$
\boxed{w(z)=\frac{\Gamma}{2\pi i}
\left[\log(z-z_1)
-\log\left(z-\frac{a^2}{\bar z_1}\right)+\log z\right].}
$$
The reflected <vortex> at $a^2/\bar z_1$ has strength $-\Gamma$, while the central term has strength $+\Gamma$. Their combined <circulation> around the cylinder is zero; this is the <circulation> normalization implicit in the requested equation.
The <vortex> moves with the <velocity> induced by the other singularities, excluding its own singular term. Since <complex velocity> is $dw/dz=u-iv$, its value at the <vortex> is $d\bar z_1/dt$. Thus
$$
\frac{d\bar z_1}{dt}
=\frac{\Gamma}{2\pi i}
\left[-\frac1{z_1-a^2/\bar z_1}+\frac1{z_1}\right]
=\boxed{\frac{i\Gamma a^2}{2\pi z_1(|z_1|^2-a^2)}.}
$$
Writing $z_1=re^{i\theta}$ and multiplying this equation by $z_1$ gives
$$
r\dot r-ir^2\dot\theta
=\frac{i\Gamma a^2}{2\pi(r^2-a^2)}.
$$
Its real part implies $\dot r=0$, and its imaginary part gives
$$
\boxed{\dot\theta=-\frac{\Gamma a^2}
{2\pi r^2(r^2-a^2)}.}
$$
The radius is constant and the angular <velocity> is constant on that orbit. For positive $\Gamma$ the orbit is clockwise with this convention. Retaining the conjugate on the time derivative is essential for the sign.
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