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

⇒ 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