In RT △ ABC, ∠ C = 90 °, AC = 8, BC = 6, P is the moving point on the inscribed circle m of △ ABC, and the minimum value of the sum of the areas of the three circles with PA, Pb and PC as diameters is obtained

In RT △ ABC, ∠ C = 90 °, AC = 8, BC = 6, P is the moving point on the inscribed circle m of △ ABC, and the minimum value of the sum of the areas of the three circles with PA, Pb and PC as diameters is obtained

Establish the coordinate system, let a (8,0), B (0,6), C (0,0), P (x, y), △ ABC inscribed circle radius be r. ∵ triangle ABC area s = 12ab × AC = 12 (AB + AC + BC) r = 24, the solution is r = 2, that is, the inscribed circle center coordinate is (2,2) ∵ P on the inscribed circle ∵ (X-2) 2 + (Y-2) 2 = 4 ∵ P point to a, B, C square sum of distance is D = x2 + Y2 + (X-8) 2 + Y2 + x2 + (y-6) 2 = 3 (X-2) 2 + 3 (Y-2) 2-4x + 76 = 88-4x, obviously 0 ≤ x ≤ 4, i.e. 72 ≤ D ≤ 88, the minimum sum of the three circles with PA, Pb and PC as diameters is 18 π