 Solution

For a circle in the right half-plane use $0<\lambda <1$. Its centre and radius from part (ii) satisfy $d^2-R^2=a^2$. Hence choosePut ${\tau }_0=\mathrm{arcosh}(d/R)$. Then $\sinh {\tau }_0=a/R$ andIndeed the Apollonius circle has centre $a\coth {\tau }_0=d$ and radius $a/\sinh {\tau }_0=R$. Its logarithmic modulus is therefore $-{\tau }_0$, while on $x=0$ the modulus is one. The logarithmic Dirichlet potential between a circle and a line isThe logarithmic singularity at $a$ lies inside the excluded disk, since $d-a<R$. The other singularity, at $-a$, lies outside the right half-plane. Thus the real logarithmic modulus is harmonic throughout the required region and has exactly the two specified boundary values.

Local branches of the complex logarithm justify harmonicity; its real part is globally single-valued even though the complex logarithm need not be single-valued around the excluded disk. The displayed solution is bounded and tends to zero at infinity. A boundedness or corresponding far-field condition supplies uniqueness in this unbounded domain; finite boundary values alone do not rule out unbounded additional harmonic solutions. The same geometry appears in cylinder-plane image charges.