I found this problems three years ago. But I never have been a proof. Recently I posted in math.stackexchange.com. I am looking for a solution of the following problems:

**A chain of six circles associated with a conic.**

Let $A_1, ..., A_6$ and $B_1, ..., B_6$ be 12 points lying on a conic, and suppose that for $i=1, ..., 5$ through $A_i, A_{i+1}, B_{i+1}, B_i$ passes a circle $(O_i)$. Then through $A_6, B_6, A_1, B_1$ as well passes a circle $(O_6)$. Let $P_1, P_4$ be intersection points of $(O_1)$ and $(O_4)$; the same for $P_2, P_5$ and $P_3, P_6$. Show that:

Three lines $O_1O_4$, $O_2O_5$, and $O_3O_6$ have a common point $O$.

Six points $P_1, ..., P_6$ lie on a circle with center in $O$.

**My remark:** With six points and six lines we get the Pascal theorem. With 12 points and six circles we have this problem

I checked it by Geogebra, But I can not calculate.

See also: Sequences of Concyclic Points on a Conic and Geogebra A chain of six circles associated with a conic