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.