As $f(x)$ is an odd function, both $f(x)$ and the circle are symmetrical with respect to the origin, we just need to ensure the circle covers at least one extreme value of $f(x)$. Let $\frac{\pi x}{k} = \frac{\pi}{2}$, we find the closest extreme point from the origin is $\big(\frac{k}{2}, \sqrt{3}\big)$. Its distance to the origin cannot be longer than $|k|$. Hence, we have $$\Big(\frac{k}{2}\Big)^2 +\Big(\sqrt{3}\Big)^2 \le k^2\implies k\in (-\infty, -2]\ \cup\ [2,+\infty)$$