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

Let O and O’ be the centres of the circles of radii 5 cm and 3 cm respectively and let PQ be their common chord.

OP = 5 cm, O’P = 3 cm, OO’ = 4 cm.

Let OL = x, so LO’ = 4 - x

Let PQ = y, so PL = y/2

In right triangle OLP,

OL^{2} = (OP^{2} – LP^{2}) = 52 – (y/2)^{2}

⇒ x^{2} = 25 – y^{2}/4 …(i)

In right triangle O’LP, O’L^{2} = (32 – y^{2}/4)

(4 – x)^{2} = 9 – y^{2}/4 …(ii)

From (i) and (ii) we get

x^{2} – 16 + 8x – x^{2} = 25 – y^{2}/4 – 9 + y^{2}/4

⇒ 8x =32

⇒ x = 4

From equation (i)

16 = 25 – y^{2}/4 => y^{2}/4 = 9 => y^{2}= 36 => y = 6

Thus, the length of the common chord is 6 cm.