Question
Two circles with centres $O$ and $O'$ intersect at two points $A$ and $B$. $A$ line $PQ$ is drawn parallel to $OO'$ through $A$ or $B$, intersecting the circles at $P$ and $Q$. Prove that $PQ = 2OO'$.
✓
Answer
Given: Two circles with centres $O$ and $O'$ intersect at two points $A$ and $B$.
Draw a line $PQ$ parallel to $OO'$ through $B$, $OX$ perpendicular to $PQ$, $O'Y$ perpendicular to $PQ$, join all.
We know that perpendicular drawn from the centre to the chord, bisects the chord.
$\therefore$ $PX = XB$ and $YQ = BY$
$\therefore$ $PX + YQ = XB + BY$
On adding $XB + BY$ on both sides,
we get $PX + YQ + XB + BY = 2(XB + BY)$
$\Rightarrow PQ = 2(XY) $
$\Rightarrow PQ = 2(OO')$
Hence, $PQ = 2OO'$
Draw a line $PQ$ parallel to $OO'$ through $B$, $OX$ perpendicular to $PQ$, $O'Y$ perpendicular to $PQ$, join all.
We know that perpendicular drawn from the centre to the chord, bisects the chord.
$\therefore$ $PX = XB$ and $YQ = BY$
$\therefore$ $PX + YQ = XB + BY$
On adding $XB + BY$ on both sides,
we get $PX + YQ + XB + BY = 2(XB + BY)$
$\Rightarrow PQ = 2(XY) $
$\Rightarrow PQ = 2(OO')$
Hence, $PQ = 2OO'$
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