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Question

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

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Solution


Let the radius of the circle centered at O and O be 5 cm and 3 cm respectively.

OA=OB=5 cm

OA=OB=3 cm

OO will be the perpendicular bisector of chord AB.

AC=CB

It is given that, OO=4 cm

Let OC be x. Therefore, OC will be x4

In OAC, we have

OA2=AC2+OC2

52=AC2+x2

25x2=AC2...(1)

In OAC, we have

OA2=AC2+OC2

32=AC2+(x4)2

9=AC2+x2+168x

AC2=x27+8x...(2)

From equations (1) and (2), we obtain

25x2=x27+8x

8x=32

x=4

Therefore, the common chord will pass through the center of the smaller circle i.e., O and hence, it will be the diameter of the smaller circle.


Length of the common chord AB=2OA=(2×3) cm=6 cm


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