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The conjecture refer the reader about the Bundle's theorem configuration. (This conjecture from a note)

Consider the Bundle theorem configuration :

Points $A_1, A_2, A_3, A_4$ lie on a circle,

points $B_1, B_2,B_3, B_4$ lie on a circle,

points $A_1, A_2, B_1, B_2$ lie on a circle,

points $B_1, B_2, A_3, A_4$ lie on a circle,

points $B_3, B_4, A_3, A_4$ lie on a circle,

and points $A_3, A_4, A_1, A_2$ lie on a circle.

Let the pair of circles $(P_1P_3Q_i)$ and $(P_2P_4Q_j)$ is such that {if $P=A$ then $Q=B$} or {if $P=B$ then $Q=A$} and if {$i=1$ then $j=2$} or {if $i=2$ then $j=1$} or {if $i=3$ then $j=4$} or {if $i=4$ then $j=3$}. Hence, there are eight pair of circles with this definition.

With one pair of circles $(P_1P_3Q_i)$ and $(P_2P_4Q_j)$ we have two common points. Hence, we have 16 points of intersection of $8$ pairs of circles.

The conjecture: The sixteen points of intersection of the eight pairs of circles lie on a circle.

The problem is true for Euclidean plane geomtry, and it be constructed on Bundle theorem configuration, the bundle theorem true for Möbius plane. I don't know the conjecture is also true for the Möbius plane? I didn't find what I'm looking for.

enter image description here

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    $\begingroup$ Do you ask if the conjecture holds for all geometries satisfieing the axioms of planar Möbius geometry (ovoidal according to the Wikipedia page you quoted)? I am not aware of the notion of a line in Möbius geometry, so I cannot make sense of Conjecture 1, and in Conjecture 2 I don't know what "with center $O$" is supposed to mean. So maybe only the slightly weaker version of Conjecture 2 without specification of a center has a chance to be proven. $\endgroup$ – Sebastian Goette Mar 2 '16 at 12:32
  • $\begingroup$ Dear Dr. @Sebastian Goette, I thank to You very much. Yes Ok, I agree with you. Could you help me give your proof of conjecture 2 without O. Thank to You very much again. $\endgroup$ – Oai Thanh Đào Mar 2 '16 at 15:06
  • $\begingroup$ I am sorry, I don't have a proof yet. I just found the problem interesting and tried to understand the question. $\endgroup$ – Sebastian Goette Mar 2 '16 at 16:32
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    $\begingroup$ Dear Dr. @SebastianGoette, I edited the question. $\endgroup$ – Oai Thanh Đào Mar 2 '16 at 17:20
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There is a circle $\omega$ perpendicular to all yellow circles (the center of it is intersection of $A_1A_2$ and $A_3A_4$). If you make inversion in $\omega$ points $A_1$ and $A_2$ go to each other, the same holds for pairs $A_3$ and $A_4$, $B_1$ and $B_2$, $B_3$ and $B_4$. That means that your circles $(P_1P_3Q_i)$ and $(P_2P_4Q_j)$ symmetric with respect to $\omega$, therefore the points of intersection of them lie on $\omega$, which is the green circle on your figure.

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