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Two circles intersecting in $A$ and $B$. They have a common tangent, with point $P$ on circle one and point $Q$ on circle two. Prove that the line through $A$ and $B$ cuts the line $PQ$ by half.

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  • $\begingroup$ Let the intersection of AB and PQ be R. Since AB is the radical axis and hence tangents from R to the two circles are of equal length and hence RP = RQ $\endgroup$ Oct 31, 2017 at 17:07

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Let $R$ be the intersection of the line $PQ$ and the line $AB$. Then by the Tangent-Secant Theorem $$|PR|^2=|RA|\cdot|RB|=|RQ|^2.$$

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Similarly in spirit to Robert Z's answer, invert about the intersection point with inversion circle passing through $P$. Then, the entire diagram looks the same under inversion, hence the distance $QR$ is the inversion radius, which is $PR$.

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Let the point of intersection of $PQ$ and $BA$ (produced) be $X$. Now, using tangent secant theorem, $$XP^2 = XA \cdot XB. \tag{1}$$

Similarly, $$XQ^2 = XA \cdot XB. \tag{2}$$

Using (1) and (2), $$XP^2 = XQ^2 \Longrightarrow XP = XQ.$$

Therefore the line through $A$ and $B$ cuts the line $PQ$ by half.

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