A classical exercise in working with "relative" figures. Draw a circle of radius $1$ for each of the $251$ points as centres; the sum of their areas is $251\pi$, and they are contained in the circle of same centre as the original one and radius $5$, having area $25\pi$. Since the sum of the areas of the "small" circles is more than $10$ times that of the "big" circle, there will exist a point covered by at least $11$ of the "small" circles. A circle of radius $1$ centred at that point will contain at least those $11$ points that were the centres of those above mentioned $11$ "small" circles