= Solution
For the upper <Apollonius circle>, let its constant ratio be $P>1$. Matching the center and radius from part (i) gives
$$
y_0=\frac{b(P^2+1)}{P^2-1},\qquad R=\frac{2bP}{P^2-1}.
$$
Subtracting their squares yields $\boxed{b=\sqrt{y_0^2-R^2}}$. Also
$$
\frac{y_0}{R}=\frac{P+P^{-1}}2=\cosh(\log P),\qquad\log P=\operatorname{arcosh}(y_0/R).
$$
The relevant <inverse hyperbolic cosine> is positive. Now $\log|(z+ib)/(z-ib)|$ is the <real part> of a local <holomorphic logarithm>, hence a <harmonic function> away from $z=\pm ib$. The point $ib$ is inside the excluded disk, because $0<y_0-b<R$, while $-ib$ is below the real axis. Thus neither singularity lies in the physical domain.
On $y=0$ the numerator and denominator have equal <modulus>, so the logarithm is zero. On the circular boundary its value is $\log P=\operatorname{arcosh}(y_0/R)$. Therefore the <harmonic potential between a grounded line and a circular boundary> is
$$
\boxed{V(x,y)=\frac{V_0}{\operatorname{arcosh}(y_0/R)}\log\left|\frac{x+iy+i\sqrt{y_0^2-R^2}}{x+iy-i\sqrt{y_0^2-R^2}}\right|.}
$$
This verifies <Laplace's equation> and both prescribed <Dirichlet boundary conditions>. It also tends to zero at infinity. A uniqueness assertion on this unbounded domain would require an additional boundedness or infinity condition; the displayed function is the natural decaying solution requested.
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