 Solution

Choose electric scalar potential $\Phi =-Ex$ and the symmetric gauge $\mathbf{A}=B(-y,x,0)/2$, whose curl is $B\hat{\mathbf{z}}$. Minimal coupling gives $L=m{\dot{\mathbf{r}}}^2/2-k{\mathbf{r}}^2/2+q\mathbf{A}\cdot \dot{\mathbf{r}}-q\Phi$. Since $x=\rho \cos \theta$ and $x\dot{y}-y\dot{x}={\rho }^2\dot{\theta }$,The magnetic term is velocity dependent; it is not an ordinary position-only potential.