It is known that the circle x2 + Y2 + x-6y + M = 0 and the line x + 2y-3 = 0 intersect at different points P and Q. if op ⊥ OQ (o is the coordinate origin), then M=______ ．

The simultaneous linear and circular equations are obtained as follows: (2y-3) 2 - (2y-3) + y2-6y + M = 0, then: Y1 + y2 = 4, Y1 · y2 = m + 125  x1 · x2 = (- 2y1 + 3) · (- 2Y2 + 3) = 4y1y2-6 (Y1 + Y2) + 9 = 4 · m + 125-15 known op ⊥ OQ, then Kop * koq = - 1, that is: Y1 · Y2 + x1 · x2 = 0 ⊥ m + 125 + 4 · m + 125-15 = 0, that is, M + 12-15 = 0 ⊥ M = 3, so the answer is: 3

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