Brief answer: yes, it's possible. Here's a somewhat scrappy answer: this is pretty much a transcription of what I wrote down when working it out, so it's a bit messy: sorry.
First of all I'll use what I think is the mathematicians' version of spherical polar coordinates (apparently physicists use the two angle names swapped). So starting from a right-handed cartesian coordinate system $(x, y, z)$, the spherical polars are $(r, \theta, \phi)$, where
$$
\begin{align}
x &= r \sin\phi \cos\theta\\
y &= r \sin\phi \sin\theta\\
z &= r \cos\phi
\end{align}
$$
(So here, $\theta$ is the angle in the $x$-$y$ plane, and $\phi$ is the angle to the $z$ axis).
So let's set up an inertial coordinate system whose origin is the centre of the Earth (so this system is not rotating with the Earth). Obviously we'll take the Earth to be perfectly spherical, its density will be a function only of $r$ and the orbit of the satellite will be perfectly circular.
The satellite will orbit about the $y$ axis, so the coordinates of the satellite, if we pick $t=0$ suitably, are
$$(r_s, 0, \omega_s t)$$
OK, so now pick an Earth-based coordinate system, which does rotate with the Earth, $(R, \Lambda, \Phi)$. Here:
$$
\begin{align}
R &= r\\
\Lambda &= \theta - \omega_E t\quad\omega_E > 0\\
\Phi &= \phi
\end{align}
$$
Again I've ignored a boring constant which defines the planet's angle at $t = 0$.
So we can project the satellite's motion down onto the surface of the Earth, and the projected path, in the Earth-based coordinates, is then
$$(R_E, -\omega_E t, \omega_s t)$$
And this all makes sense: the satellite moves in the $-\lambda$ direction over time – westward – and it moves in the positive $\Phi$ direction.
This curve is periodic if $\omega_E/\omega_s$ is rational. The simplest case of this is $\omega_E = \omega_s$, in which case the curve looks like $(R_E, -\omega t, \omega t)$.
So then the trick is to project this last curve down onto the $X$-$Y$ plane (using capital letters for Earth-centred coordinates again). Using polar coordinates $(\rho, \Lambda)$ in that plane (I'm running out of variants of '$r$' here), then
$$
\begin{align}
\rho &= R \sin\Phi\\
\Lambda &= \Lambda
\end{align}
$$
So the projection of the curve followed by the satellite on the surface of the Earth looks like $(R_E \sin\omega t, -\omega t)$. To see that this is a circle, change to cartesian coordinates $(X, Y)$:
$$
\begin{align}
X &= \rho\cos\Lambda\\
Y &= \rho\sin\Lambda
\end{align}
$$
(I confess at this point to have lost track over whether these are the same $(X, Y)$ as above, but it does not matter).
So
$$(X, Y) = R_E(\sin\omega t \cos\omega t, -\sin\omega t \cos\omega t)$$
And now we can use some trig identities:
$$
\begin{align}
\sin\theta\cos\theta &= \frac{\sin 2\theta}{2}\\
\sin\theta\sin\theta &= \frac{1}{2} - \frac{\cos 2\theta}{2}
\end{align}
$$
to get, finally,
$$(X, Y) = \frac{R_E}{2}(\sin 2\omega t, \cos 2\omega t -1)$$
Which is the equation of a circle in the $(X, Y)$ plane, radius $R_E/2$ and centre $(0, -R_E/2)$.
And that means that yes, it's Viviani's curve.
