= Solution
For a vacuum <electric potential>, put $\Phi=R(r)Y(\theta)$ in <Laplace's equation> in <spherical coordinates>. Multiplication by $r^2/(RY)$ separates the radial and angular terms. With separation constant $\ell(\ell+1)$ and $x=\cos\theta$, the angular <Legendre differential equation> is
$$
\boxed{\frac d{dx}\left[(1-x^2)\frac{dP_\ell}{dx}\right]+\ell(\ell+1)P_\ell=0.}
$$
Regularity at both polar axes selects nonnegative <integers> $\ell$ and the <Legendre polynomials>. The radial equation is $(r^2R')'-\ell(\ell+1)R=0$, whose independent solutions are $r^\ell$ and $r^{-\ell-1}$. Hence the angularly regular axisymmetric vacuum solution is
$$
\boxed{\Phi(r,\theta)=\sum_{\ell=0}^\infty(A_\ell r^\ell+B_\ell r^{-\ell-1})P_\ell(\cos\theta).}
$$
The radial <domain> and its <boundary conditions> determine which terms are admissible.
For the given <charge density>, only $P_1(\cos\theta)=\cos\theta$ is forced. Write $\Phi=f(r)\cos\theta$. The interior radial <Poisson equation> is $f''+2f'/r-2f/r^2=-4$. The identity for the <Poisson equation with a spherical harmonic source> gives the particular solution $f_p=-r^2$, since the radial operator takes $r^2$ to four. Finiteness at zero and decay at infinity therefore give
$$
\Phi_{\rm in}=(Ar-r^2)\cos\theta,\qquad\Phi_{\rm out}=Br^{-2}\cos\theta.
$$
Continuity of the <electric potential> and its radial <derivative> at $a$ requires $Aa-a^2=B/a^2$ and $A-2a=-2B/a^3$. Solving gives $A=4a/3$, $B=a^4/3$, so
$$
\boxed{\Phi(r,\theta)=\begin{cases}(\tfrac43ar-r^2)\cos\theta,&0\leq r\leq a,\\\dfrac{a^4}{3r^2}\cos\theta,&r\geq a.\end{cases}}
$$
All other angular modes have zero coefficients after imposing regularity, matching and decay. At the origin the interior expression is $(4a/3)z-rz$, which extends with a continuous <gradient> and value zero. The angular <charge density> has no unique value exactly at that point; <Poisson's equation> holds classically for $r>0$ and as a <weak solution> through the origin. There is no additional point charge.
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