= Solution
Write $\gamma=b/(2m)$ and $\nu=\sqrt{\omega^2-\gamma^2}$, where $\omega=\sqrt{k/m}$ is the undamped <angular frequency>. The <linear drag> gives the vector <damped harmonic oscillator> equation
$$
\ddot{\mathbf r}+2\gamma\dot{\mathbf r}+\omega^2\mathbf r=0.
$$
For the original circular <initial conditions>, $\mathbf r_0=(c,c)$ and $\mathbf v_0=\omega(-c,c)$, the exact <underdamped> solution is
$$
\mathbf r(t)=e^{-\gamma t}\left[\mathbf r_0\cos\nu t+\frac{\mathbf v_0+\gamma\mathbf r_0}{\nu}\sin\nu t\right].
$$
Using $|\mathbf r_0|^2=2c^2$, $|\mathbf v_0|^2=2\omega^2c^2$ and $\mathbf r_0\cdot\mathbf v_0=0$ gives
$$
\boxed{U(t)=kc^2e^{-2\gamma t}\left[1+\frac\gamma\nu\sin2\nu t+\frac{2\gamma^2}{\nu^2}\sin^2\nu t\right].}
$$
This exact <potential energy> is useful because it satisfies the prescribed <initial conditions>, including $\dot U(0)=k\mathbf r_0\cdot\mathbf v_0=0$. In the stated weak-<damping> limit $\gamma\ll\omega$, $\nu\simeq\omega$ and the square bracket differs from one by a small oscillatory term. The leading approximation is therefore
$$
\boxed{U(t)\simeq kc^2e^{-bt/m}.}
$$
The radius is approximately $\sqrt2\,c\,e^{-bt/(2m)}$, so the trajectory is approximately a shrinking <logarithmic spiral>. The exactly exponential result would require slightly adjusted initial <velocity>; with the specified undamped circular <velocity>, the small modulation above is real. This is the <initially circular damped planar oscillator>.
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