The answer is 3. Take an arbitrary point $A$ on the circle, then make points $B$ and $C$ on such circle such that $AB = AC = R\sqrt{2}$ so $BC = 2R$. Then, an entire semicircle (and an extra segment) is not available for a point to lie in. Now pick a point $D$ on the circle outside the segment. It will encapsulate a semicircle of the original circle and an extra segment also. To maximize the available remaining area, make sure these two overlap as much as possible, so for example $D$ is arbitrarily close to $B$ or $C$ on the circle. This would still mean more than three quarters of the original circle is not available to be placed a point anymore, so we can only populate the last sector and we are out of space. The language used in this "proof" might not seem rigorous enough, so I hope someone can improve my solution. Thanks!