= Solution
For <overdamping>, $\gamma=b/(2m)>\omega$. Put $\alpha=\sqrt{\gamma^2-\omega^2}>0$. Both <characteristic roots> $-\gamma\pm\alpha$ are negative real numbers. Replacing trigonometric functions by <hyperbolic functions> gives the exact trajectory
$$
\boxed{\mathbf r(t)=e^{-\gamma t}\left[\mathbf r_0\cosh\alpha t+\frac{\mathbf v_0+\gamma\mathbf r_0}{\alpha}\sinh\alpha t\right].}
$$
It starts from $(c,c)$ with tangent $(-\omega c,\omega c)$, bends anticlockwise toward the origin, and approaches it without repeated revolutions. The right panel gives a representative choice $\gamma=1.5\omega$; the paper does not specify one particular value of $b$ in this regime.
One can make the qualitative shape precise. In radial and tangential directions relative to the initial radius, the angle turned from the starting direction obeys
$$
\tan\Delta\phi(t)=\frac{\omega\sinh\alpha t}{\alpha\cosh\alpha t+\gamma\sinh\alpha t}\quad\Longrightarrow\quad\boxed{\Delta\phi(\infty)=\arctan\frac\omega{\gamma+\alpha}<\frac\pi4.}
$$
For $c>0$, both coordinates stay positive, and the approach is along the fixed ray with $y/x\to(\gamma+\alpha+\omega)/(\gamma+\alpha-\omega)$. The initial <velocity> is tangential, so an immediately straight radial return would be incorrect. The absence of oscillatory <characteristic roots> prevents repeated revolutions. This is a <non-spiralling overdamped central oscillator trajectory>.
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