= Solution
Differentiate the <Lagrangian> with respect to the generalized velocities:
$$
\boxed{p_j=\frac{\partial L}{\partial\dot x_j}=m\dot x_j+eA_j.}
$$
The <Legendre transform> cancels the vector-<potential energy> term, giving $H=p\cdot\dot x-L=m|\dot x|^2/2+V$. Eliminating <velocity> with $m\dot x=p-eA$ therefore yields
$$
\boxed{H=\frac{|p-eA|^2}{2m}+V.}
$$
For the given gauge, set $\pi_1=p_1+eBx_2/2$, $\pi_2=p_2-eBx_1/2$, $\pi_3=p_3$. <Hamilton's equations> give $\dot x_j=\pi_j/m$, $\dot p_1=(eB/2)\dot x_2$, $\dot p_2=-(eB/2)\dot x_1$, $\dot p_3=0$. Hence the conserved quantities are
$$
\boxed{p_3,\qquad K_1=p_1-\frac{eB}{2}x_2,\qquad K_2=p_2+\frac{eB}{2}x_1,\qquad H.}
$$
The two transverse combinations are components of <magnetic pseudomomentum>. \b[The printed conserved combinations omit the <electric charge> factor $e$.] With the stated <vector potential> and general <electric charge> they need the factors displayed above; as printed they hold only if $e=1$ or if <electric charge> is already absorbed into the field convention.
The initial data give $K_1=K_2=p_3=0$. Thus $p_1=eBx_2/2$, $p_2=-eBx_1/2$ throughout, and
$$
\dot x_1=\omega_cx_2,\qquad\dot x_2=-\omega_cx_1,\qquad\omega_c=eB/m.
$$
Solving with the initial point gives
$$
\boxed{x_1=a\sin(\omega_ct),\quad x_2=a\cos(\omega_ct),\quad x_3=0.}
$$
Consequently $x_1^2+x_2^2=a^2$: the <guiding centre> is the origin and the signed <cyclotron frequency> fixes the rotation direction. The <energy> is the constant $e^2B^2a^2/(2m)$.
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