Above the picture first. Convert to a plane rectangular coordinate system. Let o be the origin and D(-3√? 2)，C(√？ 2,0) According to the question, A is above the Y axis and B is below. Let ∠ADB=m, then ∠DCB=m+π/4, and let the slope of straight line AD be k 1=k, then K2 = tan (m+π/4) = (1+k)/(1. Left circle: (x+3 √ 2) 2+y 2 =1... ①? Right circle: (x-√ 2) 2+y 2 =1... ②.
Straight line ad: y = k 1 (x+3 √? 2), brought into ①: (y/k1) 2+y 2 =1,∫y is positive, ∴ y = √ (k12/(k65438+65438).
BC line: y = k2 (x-√? 2), similarly: (y/k2) 2+y 2 = 1, ∫y is positive, ∴ y = √ (k2 2/(k2 2+1))
Let A(x 1, y 1) and B(x2, y2). y 1=k 1(x 1+3√？ 2)，y2=k2(x2-√？ 2)。
∫y2-y 1 = x2-x 1 =(y2/k2)+√？ 2-y 1/k 1+3√？ 2, and y 1, y2, k 1, k2 can all be transformed into equations about k. If k has no solution, then AD is perpendicular to the X axis. . (PS: 10 is not enough. Money is out of proportion to time. Thanks =? =)