The most simple setup is to consider just a spherical conducting shell of radius ##a## around the origin. Now put a charge inside the shell. We choose the reference frame such that this charge sits at ##\vec{r}_0=r_0 \vec{e}_3##. For simplicity assume that the shell is grounded, i.e., we look for a solution of the Poisson equation
$$\Delta \phi=-q \delta^{(3)}(\vec{r}-\vec{r}_0)$$
with the boundary condition
$$\phi(\vec{r})|_{|\vec{r}|=0}=0.$$
It is now immidiately clear that in this situation the unique solution for ##|\vec{r}|>a## is the Coulomb potential fulfilling the boundary conditions since there are no charges outside and the spherical shell ##|\vec{r}|=a## is an equipotential surface. Thus we have
$$\phi(\vec{r})=\frac{q}{4 \pi r}-\frac{q}{4 \pi a}.$$
For ##|\vec{r}|<a## we use the image-charge method. By symmetry the image charge ##q'## must sit on the ##x_3## axis outside of the shell. To find ##q'## and its location ##\vec{r}_0'=r_0 \vec{e}_3## we work in spherical coordinates. Inside we have a field given by the charge ##q## and the image charge ##q'##:
$$\phi(\vec{r})=\frac{q}{4 \pi \sqrt{r^2+r_0^2-2 r r_0 \cos \vartheta}} + \frac{q'}{4 \pi \sqrt{r^2+r_0^{\prime 2} -2 r r_0' \cos \vartheta}}.$$
Now for ##r=a## this must be 0, i.e.,
$$q \sqrt{a^2+r_0^{\prime 2} -2 a r_0' \cos \vartheta}=-q' \sqrt{a^2+r_0^2-2 a r_0 \cos \vartheta}.$$
Squaring this gives
$$q^2 (a^2+r_0^{\prime 2} -2 a r_0' \cos \vartheta)=q^{\prime 2} (a^2+r_0^2-2 a r_0 \cos \vartheta).$$
This can only be true if the coefficients in front of ##\cos \vartheta## and the part independent of ##\vartheta## are equal, i.e.,
$$q^2 r_0'=q^{\prime 2} r_0$$
$$q^2 (a^2+r_0^{\prime 2})=q^{\prime 2} (a^2+r_0^2).$$
This leads to a quadratic equation for ##r_0'## which is uniquely solved under the constraint that ##r_0'>a## must hold,
$$r_0'=\frac{a^2}{r_0}$$
Plugging this into the first equation yields
$$q'=\pm \frac{q}{r_0} q.$$
Plugging it into the ansatz for our potential the boundary condition ##\phi(\vec{r})|_{|\vec{r}|=a}## gives then uniquely
$$q'=-\frac{a}{r_0} q.$$
So our potential for ##r \leq a## reads
$$\Phi(\vec{r})=\frac{q}{4 \pi \sqrt{r^2+r_0^2 - 2 r r_0 \cos \vartheta}}-\frac{q a}{4 \pi \sqrt{r^2 r_0^2+a^4 - 2 a^2 r r_0 \cos \vartheta}}.$$