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

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 ∆OAC,

OA2 = AC2 + OC2

⇒ 5 2 = AC2 + x2

⇒ 25 − x2 = AC2 ... (1)

In ∆O'AC,

O'A2 = AC2 + O'C2

⇒ 3 2 = 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 = 6m.