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Put $\kappa =qB/m$, retaining its sign. The Lorentz force equations areThus $z=0$ remains invariant when $z(0)=\dot{z}(0)=0$. Also $d|\dot{\mathbf{r}}{|}^2/dt=2\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$, since a vector is perpendicular to its cross product with the magnetic field. Choose the initial planar velocity along the positive $x$ axis and the initial location at $(0,0)$. Integration givesThis is a circle centred at $(0,-u_0/\kappa )$ with radius $u_0/|\kappa |$. The cyclotron frequency is $|qB|/m$. Equivalently, choosing this centre as the polar origin givesThe rotation is clockwise viewed from positive $z$ when $qB>0$, and reverses when $qB<0$. In the PDF hint, $\dot{r}=\ddot{r}=0$ are radial conditions with this centred origin; they do not require zero particle velocity. If $qB=0$, the limiting motion is a straight line instead.

With linear drag, put $a=\mu /m>0$. The governing equations for damped cyclotron motion becomeTaking the scalar product with the velocity gives $\frac{1}{2}d|\dot{\mathbf{r}}{|}^2/dt=-a|\dot{\mathbf{r}}{|}^2$, henceFor the planar path, let $w=\dot{x}+i\dot{y}$ and $\zeta =x+iy$. Then $\dot{w}=-(a+i\kappa )w$, so for arbitrary initial planar dataLet ${\zeta }_{\infty }={\zeta }_0+w_0/(a+i\kappa )$. The position relative to this limiting point isIts radius shrinks exponentially while its angle rotates uniformly: the path is a logarithmic spiral into ${\zeta }_{\infty }$, not a succession of circles about the original vacuum guiding centre. The weak-drag assumption is $\mu \ll |qB|$, so many revolutions occur before appreciable contraction.