Note: This is a rewrite. See the edits for previous versions and missteps.
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 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
So far my record is convenient to allow B to have some blocks with less than 5 points, we can easily bring all44 blocks up to 5 points.
Let me digress for some terminology (which I'll try to keep fairly standard). a update:design Here is my best result so fara pair $D=(V,B)$ where $V$ is a set of $v$ vertices and a $B$ is a collection of subsets called blocks There are various names for designs which satisfy additional conditions. I am leavingTwo are
Every pair of points is in my earlier constructionsexactly one common block. (Then $D$ is called a linear space)
Every pair of points is in case they inspire someoneat least one common block and every block has the same number $k$ of points. (Then $D$ is called a $(v,k,2)$-covering design and the La Jolla Repository has information about these.)
When both conditions hold, $D$ is called a Steiner System $S(2,k,v).$ Many that one encounters arise from algebraic constructions. This is perhaps due to improve themthe Streetlight effect. I would guess
I'll coin the term super-linear space for a design such that 30 blocks might be enoughevery pair of points from $V$ occur in at least one common block but the blocks may have various sizes (since I don't know a standard name.)
You have any justification for$50$ points and do not require that every pair of points is in a block but do wish that from every 5 points (element of $S$), at least one pair is in at least one block. You also want all blocks to have $5$ points. I'll find it convenient to only require that each block have at most 5 points, then one can arbitrarily enlarge blocks to size 5.
42 blocks All my constructions have this form: Split the points into 34 groups of sizes 40,5(or 2 or 3) and 5. For the largefor each group use the 40 lines oftake a projective $3$super-linear space of order 3 as 40 blocks each with 4no block having more than 6 points. Let the two small group be blocks. Any set inThen any element of $S$ has two points in the same group and they will bethose two points are in a common block. Perhaps one can do better without this restriction. I include a few other possibilities in case it inspires anyone to get a better result.
4444* blocks. Split into 4 groups of size 21,21,4 and 4. For each of the large setsgroups 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 determine a unique block.
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. Perhaps there is a way to cull out some of the lines.
Under 52 blocks If the points are partitioned into 43 groups of 13 and one of 11 (with one point allowed to livetwo fake points added 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 setsDelete $X,Y$ to improve this to 45 of size 4 points noand 6 of size 3 and one of size two in. Now tack the same block. But any set of size 2 onto a block of size 3 to get $51$ blocks (one of size 5 points has). (If $X,Y$ are in different groups it might be better, I haven't checked.) One can also get rid of at least two in the same group3 more blocks as follows: take a block abcd, delete it and suchadd b as a pair determinesfifth point to a unique block. Since we only have 4 points in each block containing c and another containing c, it seemsadd c as if the optimal number is quite a bit lowerfifth point to two blocks containing d and a and add d as a fifth point to two blocks containing a and b.
