= Circular orbit
Denote the orbit radius by $b$ here to distinguish it from observation distance; it is the quantity called $R$ in this part of the paper. Write
$$
\mathbf x(t_r)=b(\cos\omega t_r,\sin\omega t_r,0),\qquad
\mathbf a(t_r)=-b\omega^2(\cos\omega t_r,\sin\omega t_r,0).
$$
For observation at <spherical coordinates> $(\theta,\phi)$, measured from the orbit normal,
$$
\mathbf n\cdot\mathbf a=-b\omega^2\sin\theta\cos(\omega t_r-\phi),\qquad
\langle(\mathbf n\cdot\mathbf a)^2\rangle=\tfrac12b^2\omega^4\sin^2\theta.
$$
The acceleration magnitude is constant, $|\mathbf a|=b\omega^2$. Therefore the <radiation pattern of a circularly moving nonrelativistic charge> is
$$
\boxed{\frac{d\langle P\rangle_{\rm circular}}{d\Omega}
=\frac{q^2b^2\omega^4}{32\pi^2\epsilon_0c^3}(1+\cos^2\theta),\qquad
\langle P\rangle_{\rm circular}=\frac{q^2b^2\omega^4}{6\pi\epsilon_0c^3}.}
$$
Time averaging removes the azimuthal dependence. The axial intensity is twice the equatorial intensity, and no direction has zero mean intensity: the instantaneous dipole axis continually rotates. This is also the sum of the two orthogonal linear-oscillation patterns with their relative <phase quadrature>. At equal radius and linear-oscillation amplitude, the total mean <power> is twice the linear result. The calculation requires $b\omega\ll c$; relativistic <synchrotron radiation> is outside this approximation.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2019-b4-radiation-patterns.png]
{title=Normalized angular power for linear oscillation and nonrelativistic circular motion, with angle measured from the same z axis}
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