Consider two circles $C,C'$ in euclidean plane which intersect in exactly two points $Q,R$ and consider the line $QR$ through these points.

The claim is that a point point $P$ lies on the line $QR$ if and only if for any two lines $\ell,\ell'$ through $P$ with $C\cap\ell=\lbrace A,B \rbrace$ and $C'\cap\ell'=\lbrace A',B' \rbrace$ the following formula line segments hold $$\vert PA\vert\cdot \vert PB\vert=\vert PA'\vert\cdot \vert PB'\vert.$$

Unfortunatley I have no idea how to prove this claim. Could someone be so kind and help me?

Best wishes

  • 1
    $\begingroup$ Try to search for "power of a point". $\endgroup$
    – Mick
    Jun 21, 2015 at 16:04

1 Answer 1


The notion involved is that of "power of a point" with respect to a circle. Leaving aside some details that you may check in some textbook, given a circle $C$ and a point $P$ on the plane, one considers the intersections of $C$ and any line through $P$, say points $A$ and $B$. Then one proves that the product $PA \cdot PB$ is independent of the line chosen, and this is called the "power" of $P$ with respect to $C$.

So we can recast your question as the statement: for two circles $C$ and $C'$ intersecting at points $Q$ and $R$, the set of points of the plane with same power with respect to both circles is precisely line $QR$ (the usual terminology is "radical axis").

In order to prove this, I would suggest you to recast the condition $PA \cdot PB=PA' \cdot PB'$ into something about similar triangles.

  • $\begingroup$ Thank you very much. This helped me $\endgroup$
    – asd
    Jun 22, 2015 at 6:10

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