= Solution
A circular orbit has uniform angular motion, so its <Keplerian radial-velocity curve> is sinusoidal. Choose phase $\phi=(t-t_{\rm tr})/P$, positive stellar velocity away from the observer, and an edge-on circular orbit. One consistent convention is
$$
\boxed{V-V_0=-K\sin(2\pi\phi).}
$$
Transit occurs at the decreasing systemic-velocity crossing $\phi=0$; secondary eclipse is at the increasing crossing $\phi=1/2$. Extrema occur one quarter of an orbit away. Reversing the velocity sign convention reverses the crossing directions but not the conjunction phases.
For an eccentric orbit, $\nu$ is not proportional to time. Solve $\mathcal M=E-e\sin E$, then use $\tan(\nu/2)=\sqrt{(1+e)/(1-e)}\tan(E/2)$ in the stated velocity relation. Motion is fastest near periastron, so the curve has an abrupt feature there and a more prolonged change near apastron, with its shape also depending on $\varpi$. It is not merely a shifted sinusoid. An eccentric transit need not coincide with $V=V_0$, because the $e\cos\varpi$ offset remains at conjunction.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015-s30-radial-velocity-and-atmospheric-annulus.png]
{title=Circular and eccentric stellar radial velocities, and a transmission-spectrum annulus}
{description=Original calculated velocity curves use a circular transit as phase zero on the left, and periastron as phase zero for the separate eccentric example with $e=0.5$ and $\varpi=0$. The right panel illustrates the projected area added by one atmospheric scale height; its atmospheric thickness is exaggerated for visibility.}
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