To prove $EA = FB$ or that $CQ'$ is radical axis

Since $BQ=AQ$ (red segments) we have $$r_1^2+(m-x)^2 = r_2^2+(m+x)^2$$ so $$p(Q',c_1)- p(Q',c_2) = \big((m+x)^2-r_1^2\big)- \big((m-x)^2-r_2^2\big) =0$$ we see that $Q'$ has equal power with respect to both circles so it lies on a radical axis which is perpendicular to $OP$ so it is $CQ'$.

Now $C$ has also equal power to both circles so $$CB\cdot CF = CA\cdot CE$$ Since $CA= CB$ we have now $CF = CE$ and we are done. enter image description here

Tags:

Geometry

Euclidean Geometry

Geometric Transformation

Power Of The Point

Homothety

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