Question

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

Answer 1

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.
OA = OB = 5 cm
O'A = O'B = 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, O'C will be 4 − x.
In ΔOAC,
OA2 = AC2 + OC2
⇒ 52 = AC2 + x2
⇒ 25 − x2 = AC2 ... (1)
In ΔO'AC,
O'A2 = AC2 + O'C2
⇒ 32 = AC2 + (4 − x)2
⇒ 9 = AC2 + 16 + x2 − 8x
⇒ AC2 = − x2 − 7 + 8x ... (2)
From equations (1) and (2), we obtain
25 − x2 = − x2 − 7 + 8x
8x = 32
x = 4
Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

AC2 = 25 − x2 = 25 − 42 = 25 − 16 = 9
∴ AC = 3 m
Length of the common chord AB = 2 AC = (2 × 3) m = 6 m

Answer 2

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