= Solution
The <Milne-Thomson circle theorem> replaces a potential $f$ analytic inside the circle of radius $a$ by $f(z)+\overline{f(a^2/\bar z)}$ outside it, when its original singularities are outside. The circle becomes a <streamline>, the imposed flow at infinity is preserved, and no extra <circulation> is introduced.
Use the convention that positive <circulation> is counterclockwise, so a vortex potential is $\Gamma\log(z-z_1)/(2\pi i)$ and its derivative is $u-iv$. With $b=a^2/\bar z_1$, the circle construction gives
$$
\boxed{w(z)=\frac{\Gamma}{2\pi i}\{\log(z-z_1)-\log(z-b)+\log z\}}
$$
up to an additive constant and local logarithm branches. There is an image vortex of strength $-\Gamma$ at $b$ and a central vortex of strength $\Gamma$. Their zero net interior <circulation> is needed for the specified motion equation.
The original vortex moves with the regular velocity of the image flow, excluding its own singular velocity. Thus
$$
\frac{d\bar z_1}{dt}=\frac{\Gamma}{2\pi i}\left\{-\frac1{z_1-a^2/\bar z_1}+\frac1{z_1}\right\}
=\boxed{\frac{i\Gamma a^2}{2\pi z_1(|z_1|^2-a^2)}}.
$$
The conjugation in this equation matches the printed source. Write $z_1=re^{i\theta}$ and cancel $e^{-i\theta}$ to obtain
$$
\dot r-ir\dot\theta=\frac{i\Gamma a^2}{2\pi r(r^2-a^2)}.
$$
Real and imaginary parts give
$$
\boxed{\dot r=0,\qquad\dot\theta=-\frac{\Gamma a^2}{2\pi r^2(r^2-a^2)}}.
$$
The vortex moves uniformly on the circle determined by its initial radius, clockwise for positive $\Gamma$ in this convention.
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