= Solution
Substitution of the proposed sine and cosine displacements into either linearized equation gives $m\omega^2+qB\omega-k=0$. Thus
$$
\boxed{\omega_{1,2}=-\frac{qB}{2m}
\left[1\mp\sqrt{1+\frac{4mk}{q^2B^2}}\right].}
$$
These are signed angular frequencies, not two positive <frequency> magnitudes. To first order, with $\varepsilon\ll\rho_0$,
$$
x=\rho_0+\varepsilon\cos\omega t+O(\varepsilon^2/\rho_0),
\qquad y=\varepsilon\sin\omega t+O(\varepsilon^2/\rho_0).
$$
The orbit is therefore a \b[circle of radius $\varepsilon$ about $(\rho_0,0)$] at this order. The positive root runs counterclockwise; the negative root runs clockwise and has the larger magnitude. A positive charge in $B\hat{\mathbf z}$ alone circles clockwise. Here the spring permits an additional counterclockwise mode by providing inward force against its magnetic force.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1-tp1-circular-modes.png]
{title=Opposite-handed circular modes about the shifted spring equilibrium}
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