Solution: The diameters of two circles are on the same straight line. We can move the small circle to the center of the big circle, and during the movement, the two diameters always stay in a straight line. Let the center of the circle be O, let AB and the small circle be tangent to C, and connect o C, OA, ∴OC⊥AB to C. In Rt△OAC, the square difference between OA and OC is half of the square of AB.