What exactly does radical axis mean ?Maybe its too easy , but I don't know , so I am asking .

kk108 wrote: What exactly does radical axis mean ?Maybe its too easy , but I don't know , so I am asking . The radical axis of two circles is the locus of points(a straight line) at which tangents drawn to both circles have the same length.

A cute problem , we can solve this problem by invert center M and then used Monge d'Alembert on circle omega, $(ABCD)$ and image of omega

Let the tangent of $ \omega $ at $ X $ cuts $ AC, $ $ BD $ at $ Y, $ $ Z, $ respectively. Clearly, $ Y $ is the radical center of $ \omega, $ $ \odot (ACQ), $ $ \odot (ABCD) $ and $ Z $ is the radical center of $ \omega, $ $ \odot (BDP), $ $ \odot (ABCD), $ so the Gergonne point $ T $ $ \equiv $ $ QY $ $ \cap $ $ PZ $ of $ \triangle MYZ $ is the radical center of $ \omega, $ $ \odot (ACQ), $ $ \odot (BDP). $ On the other hand, it's obvious that $ M $ is the radical center of $ \odot (ACQ), $ $ \odot (BDP), $ $ \odot (ABCD), $ so we conclude that $ MT $ $ \equiv $ $ MX $ is the radical axis of $ \odot (ACQ), $ $ \odot (BDP). $

My guess for the other solution of the inverted problem posted by Anant would be moving the line l at a constant velocity. In this manner, both centers of the circle (ACQ) and (BDP) will move at a constant speed, thus the line joining two centers will always parallel to a fixed line. Then, we choose l as the line passes through the second intersection of gamma 1 and gamma 2 (not A). From here the problem is easier to solve

@above, That is a different idea i think. Essentially, what Ivan did was to showing that two points on $XM$ have equal powers in both of them, but i do not remember which ones. Your idea is more beautiful, so let's finish with it! (the degenerate case is also very interesting) In particular, when $\ell$ passes through $M$, we have $B=C=M$. In this case, we will prove that un-inverted problem, namely, In triangle $ABD$ the $B$ mixtilinear incircle touches $BA,BD$ at $P,Q$ respectively and the circumcircle at $X$. Prove that $(ABQ)$ and $(DBP)$ meet again on the line $BX$. Now we invert about $B$ with radius $\sqrt{BA\cdot BD}$ and reflect in the bisector of $\angle ABD$. We need to prove that in triangle $ABD$, the excircle opposite $B$ touches $BA,BD$ at points $P,Q$ respectively and $AD$ at point $E$ then lines $BE,AQ,DP$ are concurrent. This however, is obvious from the fact that $(B,C;E,PQ\cap BC)=-1$.

This however, is obvious from the fact that $(B,C;E,PQ\cap BC)=-1$.[/quote] How do use sets in geometry ?