Take a circle So and its diameter D Take a chain of circles

Take a circle So and its diameter D. Take a chain of circles S_1, S_2, S_3, such that circle S_1 is tangent to S_0 and is tangent to the diameter D at the center O; the circle S_2 is tangent to S_0, to D and to S_1; the circle S_3 is tangent to S_0, to D and to S_2 and so on. Let A_1, A_2, ... be the sequence of points of tangency of the circles S_1 and S_2; the circles S_2 and S_3 and so on. Prove there exists a circle S which contains all the points A_1, A_2, ...

Solution

S2 is tangent to S1 at A1   

S2 is tangent to S3 at A2

So if we consider circle S2, OA1 and OA2 are equal because they are tangents from an external point O to S2.

Similarly OA2 and OA3 are equal because they are tangents from an external point O to S3.

Simlar pattern will continue for S4, S5 and so on.

This means OA1 = OA2 = OA3 = so on

Hence A1, A2, A3 .... are part of circle S