# prove that in case of externally touching circles the radical axis is the transverse common tangent between them

Let S1 and S2 be two circles with centers o1 and o2 respectively. By definition, radical axis of two circles is the locus of the point from which the length of the two tangents are equal. In case of externally touching circles,I read that transverse common tangent is the radical axis, but how to prove it? How can we prove that AP=AQ or BR=BS in the above diagram? I tried it by congruency but triangle o1PA and o2PA are not congruent.

• Hint: the distance between $A$ and the point where the two circles meet is equal to both $AP$ and $AQ$. Can you finish the proof of this using properties of tangents? Oct 4, 2020 at 18:22
• An extra help for @player3236's hint, draw $AO_1$ and $AO_2$, and the touching point $T$. Show the right angles on the figure and apply Pythagoras. Oct 4, 2020 at 18:25
It is well-known that the radical axis is a line, that, in this case, passes through the tangency point, say $$T$$. If $$A$$ is a point on the common tangent of circles $$\omega_1$$ and $$\omega_2$$, it suffices to prove that $$\text{Power}_{\omega_1}(A)=\text{Power}_{\omega_2}(A)=\iff AO_1^2-r_1^2=AO_2^2-r_2^2$$ Which is a consequence of the Pythagorean theorem in $$\triangle AO_1T$$ and $$\triangle ATO_2$$.