Question

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between
their centres is 4 cm. Find the length of the common chord.​

Answer 1

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 is 6cm.

solution