= Solution
Write the <four-velocity> $U^\mu=\gamma(c,\mathbf v)$ and <four-acceleration> $A^\mu=dU^\mu/d\tau$. Since $U_\mu A^\mu=0$, spatial <orthogonality> $\mathbf U\cdot\mathbf A=0$ implies $U^0A^0=0$. Thus $A^0=c\,d\gamma/d\tau=0$ and \b[the <speed> is constant]. If “<acceleration>” is read as the laboratory three-acceleration instead, $\gamma\mathbf v\cdot d\mathbf v/dt=0$ immediately gives the same conclusion.
For <four-acceleration in uniform circular motion>, the <speed> is $r|\omega|<c$, so $\gamma=(1-r^2\omega^2/c^2)^{-1/2}$ is constant. Differentiating the <four-velocity> using $d/d\tau=\gamma\,d/dt$ gives
$$
\boxed{A^\mu=\left(0,-\gamma^2r\omega^2\cos\omega t,
-\gamma^2r\omega^2\sin\omega t,0\right).}
$$
The <proper acceleration> is $\sqrt{-A_\mu A^\mu}=\gamma^2r\omega^2$. The vanishing time component reflects constant <energy>; the spatial components describe the changing <momentum> direction.
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