## two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4 cm find the length of the common chord

Question

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## Answers ( )

Answer: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 is6cm

Step-by-step explanation:plz

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