= 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 choose
$$
\boxed{a=\sqrt{d^2-R^2}.}
$$
Put $\tau_0=\operatorname{arcosh}(d/R)$. Then $\sinh\tau_0=a/R$ and
$$
\lambda=e^{-\tau_0}=\frac{d-a}{R}=\frac{R}{d+a}.
$$
Indeed 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> is
$$
\boxed{V(x,y)=\frac{V_0}{\operatorname{arcosh}(d/R)}\log\left|\frac{z-a}{z+a}\right|.}
$$
The 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>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-q7-circle-line-potential.png]
{title=Apollonius circles and a bounded harmonic potential outside a circle beside a line}
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