Charcoal, 75 bytes
≔⊕ＮθＦθＦιＦ⌕ＡＸ…¹θ²ΣＸ⊕⟦κι⟧²⊞υ⊕⟦κιλ⟧ＦυＦυ¿∧⁼⌈ι⌈κ���ικＦ…ι²Ｅօ겋⁺λμθ⭆¹⟦↔⁻λμ⁺λμ⌈ι
Try it online! Link is to verbose version of code. Explanation: Possibly not the shortest approach, but it only uses integer arithmetic. Points B and C on the circle must have integer coordinates, along with both distances MB and MC being the integer r, therefore taking two different Pythagorean triples with the same hypotenuse will generate four solutions (although some of them may result in m > X which must then be excluded).
≔⊕Ｎθ
Increment X as this saves bytes on the ranges and conditions.
ＦθＦιＦ⌕ＡＸ…¹θ²ΣＸ⊕⟦κι⟧²⊞υ⊕⟦κιλ⟧
Loop over all p and q values to find Pythagorean triples p, q, r with p < q < r <= X.
ＦυＦυ¿∧⁼⌈ι⌈κ‹ικ
Loop over (distinct) pairs of Pythagorean triples p, q, r and p', q', r where p < p' < q' < q < r.
Ｆ…ι²Ｅօ겋⁺λμθ⭆¹⟦↔⁻λμ⁺λμ⌈ι
Loop s over p and q and s' over p' and q' where s + s' <= X and output the triple m = |s - s|', n = s + s', r.