= Solution
The outward normal of the exterior region points \b[into the excluded disc], so it is $-\mathbf r$ on the unit circle. Differentiating the <Dirichlet Green function> from (e), and writing $s=|\mathbf r'|$ and $d=|\mathbf r-\mathbf r'|$, gives
$$
\begin{aligned}
\frac{\partial G}{\partial n}
&=\frac1{2\pi}\left[-\frac{1-\mathbf r\cdot\mathbf r'}{d^2}+\frac{1-\mathbf r\cdot\mathbf p}{|\mathbf r-\mathbf p|^2}\right]\\
&=\frac{s^2-1}{2\pi|\mathbf r-\mathbf r'|^2}.
\end{aligned}
$$
Here $|\mathbf r-\mathbf p|^2=d^2/s^2$ was used. The concentrated boundary data have unit integral, so the boundary representation evaluates this exterior <Poisson kernel> at $\mathbf r=(-1,0)$. Thus the bounded solution on either part of the exterior $x$-axis is
$$
\boxed{\Phi(x',0)=\frac{x'^2-1}{2\pi(x'+1)^2}=\frac{x'-1}{2\pi(x'+1)},\qquad |x'|>1.}
$$
It is positive on both intervals, diverges as $x'\to-1^-$ at the boundary source, and tends to $1/(2\pi)$ as $|x'|\to\infty$. Boundedness at infinity selects this solution; otherwise one could add an arbitrary multiple of $\log|\mathbf r'|$ without changing the circle's boundary data.
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