*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

*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

*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