= Solution
Keep the prescribed orientation $\alpha$ and strength $m$ fixed, and now let the doublet position $b(t)$ be complex. The <Milne-Thomson circle theorem> applies instantaneously with
$$
W(z,t)=\frac{M}{z-b}+\frac{\bar M z}{a^2-\bar b z},\qquad M=me^{i\alpha}.
$$
When taking its spatial derivative, $b(t)$ is a parameter. Removing the doublet's own singular term leaves the regular <complex velocity> at its position:
$$
q_{\rm reg}(b)
=\left.\frac{\bar M a^2}{(a^2-\bar b z)^2}\right|_{z=b}
=\frac{\bar M a^2}{(|b|^2-a^2)^2}.
$$
Because $q=u-iv$, the stipulated <image-induced motion of a planar doublet> obeys $\dot b=\overline{q_{\rm reg}(b)}$, not $q_{\rm reg}(b)$ itself. Hence
$$
\boxed{\dot b=\frac{ma^2e^{i\alpha}}{(|b|^2-a^2)^2}}.
$$
Its direction is constant. Indeed $\operatorname{Im}(be^{-i\alpha})$ is constant and, since $b(0)=b_0>a$ is real, equals $-b_0\sin\alpha$. Writing $b=re^{i\theta}$ in <polar coordinates> gives the path equation
$$
\boxed{r\sin(\theta-\alpha)=-b_0\sin\alpha},
\qquad
\boxed{r(\theta)=\frac{b_0\sin\alpha}{\sin(\alpha-\theta)}}
$$
where the quotient is used when $\sin\alpha\ne0$ and the branch through $(r,\theta)=(b_0,0)$ with $r>a$ is selected. It is the straight line through $b_0$ parallel to $e^{i\alpha}$. An unambiguous parametrization, also covering the degenerate polar cases, is
$$
b=b_0+s e^{i\alpha},\qquad
\dot s=\frac{ma^2}{(|b_0+s e^{i\alpha}|^2-a^2)^2},\qquad s(0)=0.
$$
If $\sin\alpha=0$, the path is radial along the positive real axis, with $\theta=0$ and changing $r$; it cannot be represented as a single-valued $r(\theta)$. The sign of $m$ selects which direction along the line is followed. These paths apply only while the singularity remains outside the obstacle; the model prescribes no orientation evolution.
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