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
Two ciecles with centre $O$ 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 $Q.$
Draw $\text{OC}\bot\text{PA}$ and $\text{O'D}\bot\text{AQ}.$
We have to prove that $PQ = 2OO'.$
Since perpendicular from the centre to a chord bisect the chord,
so $PA = 2CA ....(1)$ and $AQ = 2AD ...(2)$ Adding $(1)$ and $(2),$
we get $PA + AQ = 2CA + 2AD $
$⇒ PQ = 2(CA + AD) = 2CD$
Hence, $PQ = 2OO'$
[$\because CD$ and $OO'$ are opposite sides of a rectangle]
Draw $\text{OC}\bot\text{PA}$ and $\text{O'D}\bot\text{AQ}.$
We have to prove that $PQ = 2OO'.$
Since perpendicular from the centre to a chord bisect the chord,
so $PA = 2CA ....(1)$ and $AQ = 2AD ...(2)$ Adding $(1)$ and $(2),$
we get $PA + AQ = 2CA + 2AD $
$⇒ PQ = 2(CA + AD) = 2CD$
Hence, $PQ = 2OO'$
[$\because CD$ and $OO'$ are opposite sides of a rectangle]
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