= Solution
The parameters give
$$
\omega=\sqrt{3.95/0.10}=6.28490\,\mathrm{rad\,s^{-1}},\qquad\gamma=\frac{0.05}{2(0.10)}=0.2500\,\mathrm{s^{-1}},\qquad\nu=6.27993\,\mathrm{rad\,s^{-1}}.
$$
The coordinate expressions for the <initially circular damped planar oscillator> are
$$
\boxed{\begin{aligned}
\frac{x(t)}c&=e^{-\gamma t}\left[\cos\nu t+\frac{\gamma-\omega}{\nu}\sin\nu t\right],\\
\frac{y(t)}c&=e^{-\gamma t}\left[\cos\nu t+\frac{\gamma+\omega}{\nu}\sin\nu t\right].
\end{aligned}}
$$
The damped <oscillation period> $2\pi/\nu=1.00052\,\mathrm s$ gives almost four anticlockwise turns in four seconds. The radius envelope falls by $e^{-\gamma(4\,\mathrm s)}=e^{-1}=0.3679$. Substitution into the exact coordinates gives
$$
\boxed{(x(4\,\mathrm s),y(4\,\mathrm s))=(0.37245c,0.36286c),\qquad r(4\,\mathrm s)/r_0=0.36769.}
$$
The left panel annotates the initial tangent, time marks and endpoint. Because $c$ is an unspecified starting coordinate, the axes are normalized by $c$ rather than assigned numerical lengths.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021-q8-damped-orbits.png]
{title=Exact <underdamped> trajectory over four seconds and a representative <overdamped> trajectory for the same initial circular <velocity>}
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