You are not using the positions at all. You have 50 points. $S$ is the set of all $\binom{50}{5}=2118760$ selections of 5 points. You want a subset $B \subset S$ such that any $s \in S$ intersects at least one $b \in B$ in at least 2 points. That is an interesting problem and does not immediately strike me as familiar. If you wanted every pair to be in a block then you would need at least $130$ blocks (since every point is in 49 pairs and every block uses up 4 of those pairs).

I could wonder why $50$ but I won't. Call a member of $B$ a **block**. A given block intersects $152026$ members of $S$ in 2 or more points. This gives a lower bound of $14$ for the possible size of $B$. Of course this a weak bound since two disjoint blocks determine $200$ members of $S$ intersecting one in 2 points and the other in 3 and another $4000$ intersecting each in 2 points. If my calculations are correct, that raises the lower bound to at least $19$ blocks. I doubt that $20$ would suffice. It is convenient to allow B to have some blocks with less than 5 points,  we can easily  bring all blocks up to 5 points.

**update:** Here is my best result so far. I am leaving in my earlier constructions in case they inspire someone to improve them.

**44** blocks. Split into 4 groups of size 21,21,4 and 4. For each of the large sets take the 21 lines of a projective plane of order 4 and let the two small groups be 4 point blocks. Now any set in $S$ has at least two points in a common group and those two pints are in a common block. I don't think that this is very near optimal but it is less blatant then the two below.

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**60** blocks: split the points into two sets of 25 and use the 30 lines of an affine plane of order 5 on each. This is overkill because any 3 element set has two points in some block. 

**52** blocks  If the points are partitioned into 4 groups of 13 (with one point allowed to live in 3 groups) and then (for each group) the lines of a projective plane of order 3 were taken as blocks this would give a solution with **52** blocks of size 4. Now there are sets of 4 points no two in the same block. But any set of 5 points has at least two in the same group and such a pair determines a unique block. Since we only have 4 points in each block, it seems as if the optimal number is quite a bit lower.