= Solution
Write the <electromagnetic field tensor> as $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$. Substitution into the <Covariant Maxwell equation with the minus-plus-plus-plus metric> gives
$$
\partial_\mu\partial^\mu A^\nu-\partial^\nu(\partial_\mu A^\mu)=-\mu_0J^\nu.
$$
Choose the <Lorenz gauge> $\partial_\mu A^\mu=0$. Since $\partial_\mu\partial^\mu=\nabla^2-c^{-2}\partial_t^2$, the <electromagnetic four-potential> obeys
$$
\boxed{\left(\nabla^2-\frac1{c^2}\frac{\partial^2}{\partial t^2}\right)A^\mu=-\mu_0J^\mu.}
$$
Convolution with the outgoing <retarded Green function> gives the <retarded electromagnetic potential>
$$
\boxed{A^\mu(t,\mathbf x)=\frac{\mu_0}{4\pi}
\int\frac{J^\mu(t_{\rm ret},\mathbf x')}{|\mathbf x-\mathbf x'|}\,d^3x',
\qquad
t_{\rm ret}=t-\frac{|\mathbf x-\mathbf x'|}{c}.}
$$
For the point particle, $\tau_*(x)$ is its unique retarded <proper time>, characterized by
$$
\boxed{R^\mu(\tau_*)R_\mu(\tau_*)=0,\qquad R^0(\tau_*)>0.}
$$
Thus $y(\tau_*)$ is where the particle's <worldline> intersects the <past light cone> of the observation event $x$; the displayed expression is the covariant form of the <Liénard–Wiechert potentials>.
On the $z$-axis put $D=\sqrt{R^2+z^2}$. The <retarded time> is $t_*=t-D/c$. For the circular orbit, the spatial velocity at that time is
$$
\mathbf v(t_*)=R\omega\bigl(-\sin(\omega t_*),\cos(\omega t_*),0\bigr).
$$
Because the displacement from the retarded source point to the observation point is orthogonal to this velocity, $|R^\nu\dot y_\nu|=\gamma cD$, while $\dot y^\mu=\gamma(c,\mathbf v)$. The factors of $\gamma$ cancel, leaving
$$
\boxed{A^\mu(t,0,0,z)=\frac{\mu_0q}{4\pi D}
\left(c,-R\omega\sin\!\left[\omega\left(t-\frac Dc\right)\right],
R\omega\cos\!\left[\omega\left(t-\frac Dc\right)\right],0\right).}
$$
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