Grégoire Nicollier
November 6, 2016

This is a by-product of the results proven in the reference below. Move point A of the figure!

6 November 2016, Created with GeoGebra

The inner and the outer circles of the figure are centered at O. They are both tangent to two other circles: the inner circle lies outside the first and inside the second. A line through the (adequately chosen) intersection I of these two other circles cuts them a second time at A and B, respectively. Show that OA=OB for any line through I. (Notice the particular collinearities when A and B are both on the inner or on the outer circle.)

Proof with complex numbers. Without loss of generality, the inner circle is the unit circle, the circle \mathcal{C}_A of radius r is centered at C_A=(1+r)e^{i\varphi} and the circle \mathcal{C}_B of radius 1+r at C_B=re^{i\psi}. The desired intersection of \mathcal{C}_A and \mathcal{C}_B is

I=(1+r)e^{i\varphi}+re^{i\psi},
clearly on both circles. Point A of \mathcal{C}_A can be written as
A=(1+r)e^{i\varphi}+re^{i\alpha}.
Point B of \mathcal{C}_B is then
B=re^{i\psi}+(1+r)e^{i(\alpha+\psi-\varphi)}
as \overrightarrow{IA}, parallel to e^{i\alpha}-e^{i\psi}, and \overrightarrow{IB}, parallel to e^{i(\alpha+\psi-\varphi)}-e^{i\varphi}, are both perpendicular to e^{i(\alpha+\psi)/2}. The segments OA and OB are congruent as they are diagonals of parallelograms with sides r and 1+r enclosing the angle \left|\varphi-\alpha\right| modulo \pi (or simply as e^{i\alpha}\overline A=e^{-i\psi}B).

Synthetic proof. Let \mathcal{C}_A and \mathcal{C}_B be centered at C_A and C_B and tangent to the inner unit circle at T_A and T_B, respectively. OC_AIC_B is a parallelogram. The isosceles triangles C_AIT_A and OT_AT_B have collinear sides C_AT_A, T_AO and parallel sides C_AI, T_BO: I lies thus on the line T_AT_B. When the line rotates by \omega about I from the position IT_AT_B, A and B rotate by 2\omega about C_A and C_B, respectively, by the inscribed angle theorem: hence the angles OC_AA and OC_BB are equal for all positions of the line. As the triangles OC_AA and OC_BB have each two sides of length r and 1+r enclosing the same angle, their third sides OA and OB are equal. A similar proof uses the tangency points to the outer circle.

Reference
 G. Nicollier, Two Six-Circle Theorems for Cyclic Pentagons, Forum Geom. 16 (2016) 347-354. 
http://forumgeom.fau.edu/FG2016volume16/FG201644.pdf