= Solution
The <image charge for a grounded conducting sphere> is obtained by the <Kelvin transform>:
$$
\boxed{q^\star=-\frac{a}{r_0}q,\qquad \mathbf r^\star=\frac{a^2}{r_0^2}\mathbf r_0=\left(0,0,\frac{a^2}{r_0}\right).}
$$
It lies inside the <electrical conductor>, outside the region where it represents the physical <electric field>. The <electric potential> is
$$
\boxed{\phi(\mathbf r)=\begin{cases}
0,&r\leq a,\\[2pt]
\displaystyle\frac1{4\pi\epsilon_0}\left(\frac q{|\mathbf r-\mathbf r_0|}-\frac{qa/r_0}{|\mathbf r-\mathbf r^\star|}\right),&r>a,\quad\mathbf r\ne\mathbf r_0.
\end{cases}}
$$
At the physical <point charge>, the <electric potential> has its usual <Coulomb's law> singularity. There is no physical singularity at the <image charge>: the interior expression is identically zero.
For $|\mathbf r|=a$, direct expansion of squared distances gives
$$
|\mathbf r-\mathbf r^\star|^2
=a^2+\frac{a^4}{r_0^2}-\frac{2a^2}{r_0^2}\mathbf r\cdot\mathbf r_0
=\frac{a^2}{r_0^2}|\mathbf r-\mathbf r_0|^2.
$$
Consequently the two terms cancel exactly, proving the zero <Dirichlet boundary condition> on the sphere. At infinity each term is $O(r^{-1})$, and more explicitly
$$
\phi(\mathbf r)=\frac{q(1-a/r_0)}{4\pi\epsilon_0r}+O(r^{-2})\longrightarrow0.
$$
Thus the <Dirichlet boundary condition> at infinity also holds. The candidate solves the <Poisson equation> with the one physical <point charge>; <uniqueness of Poisson equation>, with these <Dirichlet boundary conditions>, identifies it as the required <electric potential>.
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