 Solution

In Cartesian coordinates the Lagrangian isIntroduce a Lagrange multiplier $\lambda$ for $x^2+y^2+z^2=c^{-2}$. The Euler-Lagrange equations becomeorDifferentiating the constraint twice givesTaking the scalar product of the vector equation with $\mathbf{x}$ therefore yieldsSubstitution gives

For $c=0$, the equations have the first integralsWriting $X=x+B$ and $Y=y-A$ gives $\dot{X}=-Y$ and $\dot{Y}=X$. Hence every trajectory isThese are helices of unit angular frequency about a line parallel to the $z$ axis, including circles when $C=0$ and a straight line when $R=0$.