Another chain of six circles

I found a conjecture: A chain of six circles associated with six points on a circle (in Mobius plane).

This is a generalization of the last my previous question in Three chains of six circles. (Noting that, in this configuration: $P'_1P_1P_4P'_4$ don't lie on a circle) I am looking for a proof of the conjecture as following:

Conjecture:

In the Möbius plane, let a chain of six circles $(C_1)$, $(C_2)$, $(C_3)$, $(C_4)$, $(C_5)$, $(C_6)$, such that two neighbors circles $(C_i)$ meets $(C_{i+1})$ at two points $P_i, P'_i$ where $i=1, 2, 3, 4, 5, 6$; Such that six points $P_1, P_2, P_3, P_4, P_5, P_6$ lie on a circle.

Let $A_1$ be a point on the circle $(C_6)$, let the circle $(A_1P_1P_2)$ meets the circle $(C_2)$ at $A_2$, Let the circle $(A_2P'_3P'_4)$ meets the circle $(C_4)$ at $A_3$, Let the circle $(A_3P_5P_6)$ meets the circle $(C_6)$ at $A_4$, Let the circle $(A_4P'_1P'_2)$ meets the circle $(C_2)$ at $A_5$, Let the circle $(A_5P_3P_4)$ meets the circle $(C_4)$ at $A_6$. Then show that:

• Four points $A_6, P_5, P_6, A_1$ lie on a circle.

• Six points $A_1, A_2, A_3, A_4, A_5, A_6$ lie on a circle

The key is Miquel's theorem.

It is enough to show that the points $A_1, A_2, A_3, A_4$ lie on a common circle. Denote by $(P)$ the blue circle through the points $P_1,..., P_6$. Look at the circles $(P), (C_2), (C_3), (C_4), (C_6), (A_2P'_3P'_4A_3), (A_1P_1P_2A_2), (A_3P_5P_6A_4)$. First, the five circles $(P), (C_2), (C_3), (C_4), (A_2P'_3P'_4A_3)$ satisfy the assumptions of Miquel's theorem. Therefore the four points $A_2, P_2, P_5, A_3$ lie on a common circle $(A_2P_2P_5A_3)$. Second, the five circles $(A_2P_2P_5A_3), (A_1P_1P_2A_2), (A_3P_5P_6A_4), (P), (C_6)$ also satisfy the assumptions of Miquel's theorem. Therefore, the four points $A_1, A_2, A_3, A_4$ lie on a common circle we call $(A).$

Now keep applying the same arguments in order to show that $A_2, A_3, A_4, A_5$ lie on a common circle, which has to be $(A) $ again as the current group of four points shares with the preceding group of four points exactly three points which determine the circle $(A)$. Continue the same way in order to show that $A_3, A_4, A_5, A_6$ lie on $(A)$. Analogously $A_4, A_5, A_6, A_7$ lie on $(A)$ too. But $A_7$ lies on $(C_6)$ too, so it has to be one of the two intersection points of $(A)$ and $(C_6)$, which are $A_1$ and $A_4$. Since it cannot be $A_4$, it has to be $A_1$, that is $A_7=A_1$.