How to prove a minimum number of points inside a circle inside a square with given lengths? 
Inside a square with the side length 2017 are 10000 points.
a) Prove that there is a circle of diameter 100 in its interior
  are at least 12 of these points.
b) Prove that there is even a circle of diameter 100 in its interior are
  at least 15 points.

Attempt
square: 
a = 2017
A = 2017^2 = 4068289 area
4068289 area = 10000 points
circle: 
d = 100
r = d / 2 = 50
A(circle) =  $\pi$ $\times$ r² = 7853.98
Taken from square: 
4068289 area = 10000 points 
1 area = ${10000\over 4068289}$ points
Using circle area:
A(circle) = 7853.98 = ${78539800\over 4068289}$ points = 19.305 points
q.e.d.?
Is this a possible solution for exercise a) and b) or did I miss something?
 A: HINT:
The way I would approach it: cover the square with an arrangement of circles of diameter $100$. We want as few as possible to cover the square. So we should use an economical patching. Perhaps arranging the centers of the circles in the vertices of a lattice formed by equilateral triangles? That seems good, not too much overlap. See how many you would need. Working on paper ( or with Geogebra - they even have triangular lattices I think) is helpful. Using the box principle you could get some estimate, maybe even the solution. 
Try to cover the square with a part of the honeycomb lattice with side $50$. See how many hexagons you need to fully cover the square. 

Let's see approximately how many cells we will need to cover the square. The area of a cell is $6 \frac{\sqrt{3}}{4} \cdot 50^2$. The area of the big square is $2017^2$. So we will need approximately 
$$\frac{2017^2}{3750 \sqrt{3}}= 626. \, \ldots $$ and this estimate is low. Now let's say we can cover the square not with $627$, not even with $640$, or $650$; let's say we can cover it with $714$ hexagons. Note that
$$\frac{10 000}{714} =14.005\ldots $$ So there must be at least $15$ points inside some hexagon. 
