Two circles with centres $O$ and $O^{'}$ intersect at two points $A$ and $B$ . $A$ line $P Q$ is drawn parallel to $O O^{'}$ through $A$ $(o r B)$ intersecting the circles at $P$ and $Q$ Prove that $P Q = 2 O O^{'}$

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Solution

Firstly draw two circles with center $O$ and $O^{'}$ such that they intersect at $A$ and $B$ .

Draw a line $P Q$ parallel to $O O^{'}$ .

In the circle with center $O$ , we have:

$O P$ and $O B$ are the radii of the circle. $P B$ is the chord with $O M$ as its perpendicular bisector.

i.e. $B M = M P . . . . (1)$

In the circle with center $O^{'}$ , we have:

$O^{'} B$ and $O^{'} Q$ are the radii of the circle. $B Q$ is the chord with $O^{'} N$ as its perpendicular bisector.

i.e. $B N = N Q . . . . (1)$ $B M = M P . . . . (1)$

From (1) and (2), we have:

$B M + B N = M P + N Q \Rightarrow (B M + B N) + (B M + B N) = (B M + B N) + (M P + N Q) \Rightarrow 2 (B M + B N) = (B M + B N) + (M P + N Q) \Rightarrow 2 (O O^{'}) = (B M + M P) + (B N + N Q) \Rightarrow 2 (O O^{'}) = B P + B Q \Rightarrow 2 O O^{'} = P Q$

Hence, proved.

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