The result as follows from special configuration of merge Nine Circle Theorem and Eight Circle theorem but it is new:

Problem: Let three circle $(A)$, $(B)$, $(C)$ , let $A_c$ be arbitrary point in the circle $(C)$. Construct the circle $(O_1)$ through $A_c$ tangent to $(C)$ and $(B)$, denote $(O_1)$ tangent to $(B)$ at point $A_b$; Construct the circle $(O_2)$ through $A_b$ tangent to $(B)$ and $(A)$, denote $(O_2)$ tangent to $(A)$ at point $B_a$; Construct the circle $(O_3)$ through $B_a$ tangent to $(A)$ and $(C)$, denote $(O_3)$ tangent to $(C)$ at point $B_c$; Construct the circle $(O_4)$ through $B_c$ tangent to $(C)$ and $(B)$, denote $(O_4)$ tangent to $(B)$ at point $C_b$; Construct the circle $(O_5)$ through $C_b$ tangent to $(B)$ and $(A)$, $(O_5)$ tangent to $(A)$ at point $C_a$; Construct the circle $(O_6)$ through $C_a$ tangent to $(A)$ and $(C)$, then

1. By the Nine Circle Theorem $O_6$ also tangent to $(O_1)$ but new: the point of tangency is $A_c$.
2. Six points $A_c, A_b, B_a, B_c, C_b, C_a $ lie on a circle.

Remarks: This result 2. show that this configuration is special configuration of Eight Circle theorem, when two big circles coincide, please see Eight Circle theorem

3. Three lines $O_1O_4, O_2O_5, O_3O_6$ are concurrent;
4. Three lines $AO_1, BO_3, CO_5$ are concurrent
5. Three line $AO_2, BO_4, CO_6$ are concurrent
6. The points of concurrence in (3), (4), (5) are collinear.

Question: I am looking for a proof the result above