two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centre is 4cm. Find the length of the common chord.
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Let the common chord be AB and P and Q be the centers of the two circles.
∴AP=5cm and AQ=3cm.
PQ=4cm ....given
Now, segPQ⊥chord AB
∴AR=RB=
2
1
AB ....perpendicular from center to the chord, bisects the chord
Let PR=xcm, so RQ=(4−x)cm
In △ARP,
AP
2
=AR
2
+PR
2
AR
2
=5
2
−x
2
...(1)
In △ARQ,
AQ
2
=AR
2
+QR
2
AR
2
=3
2
−(4−x)
2
...(2)
∴5
2
−x
2
=3
2
−(4−x)
2
....from (1) & (2)
25−x
2
=9−(16−8x+x
2
)
25−x
2
=−7+8x−x
2
32=8x
∴x=4
Substitute in eq(1) we get,
AR
2
=25−16=9
∴AR=3cm.
∴AB=2×AR=2×3
∴AB=6cm.
So, length of common chord AB is 6cm.
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