Solution
= Solution
The stated condition is equivalent to $(V_0/\omega)^2=X_0^2$. Therefore the solution from (b) obeys
$$
x(t)^2+y(t)^2=X_0^2(\cos^2\omega t+\sin^2\omega t)=X_0^2.
$$
For nonzero initial data this is a \b[<circle> centred at the origin with radius $|X_0|$]. The sign of $X_0V_0$ chooses its direction of traversal. If $X_0=V_0=0$, the same algebra gives only the stationary origin, a degenerate radius-zero case rather than a genuine circular orbit.