= Solution
Put $\kappa=qB/m$, retaining its sign. The <Lorentz force> equations are
$$
\ddot x=\kappa\dot y,\qquad \ddot y=-\kappa\dot x,\qquad \ddot z=0.
$$
Thus $z=0$ 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 gives
$$
\dot x=u_0\cos\kappa t,\quad\dot y=-u_0\sin\kappa t,\qquad
x=\frac{u_0}{\kappa}\sin\kappa t,\quad y=\frac{u_0}{\kappa}(\cos\kappa t-1).
$$
This 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 gives
$$
\boxed{r=\frac{mu_0}{|qB|},\qquad \dot\theta=-\frac{qB}{m},\qquad z=0.}
$$
The 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> become
$$
\ddot x=\kappa\dot y-a\dot x,\qquad
\ddot y=-\kappa\dot x-a\dot y,\qquad \ddot z=-a\dot z.
$$
Taking the scalar product with the velocity gives $\tfrac12d|\dot{\mathbf r}|^2/dt=-a|\dot{\mathbf r}|^2$, hence
$$
\boxed{|\dot{\mathbf r}(t)|=u_0e^{-\mu t/m}.}
$$
For 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 data
$$
w=w_0e^{-(a+i\kappa)t},\qquad
\zeta=\zeta_0+\frac{w_0}{a+i\kappa}\bigl(1-e^{-(a+i\kappa)t}\bigr).
$$
Let $\zeta_\infty=\zeta_0+w_0/(a+i\kappa)$. The position relative to this limiting point is
$$
\boxed{\zeta-\zeta_\infty=-\frac{w_0}{a+i\kappa}e^{-at}e^{-i\kappa t}.}
$$
Its 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-4-cyclotron-paths.png]
{title=Undamped circular and weakly damped logarithmic-spiral charged-particle paths in a uniform magnetic field}
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