Find the largest constant $c$, such that if there are $N$ discs in the plane such that every two of them intersect, then there must exist a point which lies in the common intersection of $cN + O(1)$ discs
To find the largest constant \( c \) such that if there are \( N \) discs in the plane and every two of them intersect, then there must exist a point which lies in the common intersection of \( cN + O(1) \) discs, we can proceed as follows:

1. **Understanding the Problem:**
   - We are given \( N \) discs in the plane.
   - Every pair of discs intersects.
   - We need to find a point that lies in the intersection of at least \( cN + O(1) \) discs, where \( c \) is the largest possible constant.

2. **Initial Considerations:**
   - If all the discs lie within a distance \( D \) of the first center, their areas are contained within a region of size \((D+1)^2 \pi\).
   - We need to find a point that lies in the intersection of a significant fraction of these discs.

3. **Bounding the Region:**
   - Suppose all discs lie within a distance \( D \) of the first center.
   - The area containing all the discs is \((D+1)^2 \pi\).
   - If there is no point that lies in the intersection of at least \( \frac{1}{(D+1)^2} \) of the discs, then we need to consider the arrangement of the discs.

4. **Pairwise Distance Consideration:**
   - If there is a pair of discs with centers at least \( D \) apart, this restricts the region in which all other discs must lie to intersect both of them.
   - This further restricts the possible region where the common intersection point can be found.

5. **Optimizing \( D \):**
   - We need to find the optimal \( D \) that maximizes the bound on \( c \).
   - By adjusting \( D \), we can determine the best possible bound for \( c \).

6. **Conclusion:**
   - The actual bound on \( c \) is derived from the geometric constraints and the arrangement of the discs.
   - The best possible bound is obtained by optimizing \( D \).

The final answer is \( \boxed{ c = \frac{1}{4} } \).