Proof
Since all extended lines pass through Ñ, L and M cannot both be extended lines.
Suppose that L is an extended line. Then M must be a circle. Let it have
centre P and radius r.
If A and B are inverse with respect to L, they must lie on a line perpendicular to L.
Also, they must lie on opposite sides of L. Say that A lies on the same side as M.
If A and B are inverse with respect to M, they must lie on a line through P,
and on the same side as P.
Also one must lie inside M. Since B lies on the opposite side of L,
A is inside M.
Then L, M, A, B and P are as in the picture on the right,
and Q is the foot of the perpendicular from P to L.
Suppose that A and B are inverse with respect to L, so QA = QB.
Let N denote the circle on AB as diameter, and suppose it meets M at R.
Then QA = QB = QR.
Now, A and B are inverse with respect to M if and only if PA.PB = r^{2} = PR^{2}.
i.e. PR^{2} = PA.PB = (PQQA).(PQ+QB) = PQ^{2}QR^{2} (as QA = QB = QR).
Thus, A and B are common inverses if and only if <PRQ is a right angle.
This occurs if and only if R is a point where the circle K on PQ as diameter meets M.
Hence, there is exactly one pair {A,B}.
Now the case where L and M are circles.
We simply choose a point T on L and a circle C with centre T.
Inversion in C maps L to an extended line, as is on L (Inversion Theorem).
It send M to a circle, as T is not on M.
Since inversion maps inverses to inverses (Algebraic Inversion Theorem),
the result follows.

