Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 12 / c

The stated condition is equivalent to $(V_0/\omega {)}^2=X_0^2$. Therefore the solution from (b) obeysFor nonzero initial data this is a circle centred at the origin with radius $|X_0|$. The sign of $X_0{V}_0$ 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.