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

The diagram is shown below:

OO' will be the perpendicular bisector of chord AB

⇒ AC = CB

It is given that, OO' = 4 cm (The distance between their centres)

Let OC be x. Therefore, O'C will be 4 − x

In ΔOAC,

OA² = AC² + OC²

⇒ 5² = AC² + x²

⇒ 25 – x² = AC² ..........(i)

Now, In ΔO'AC,

O'A² = AC² + O'C²

⇒ 3² = AC² + (4 − x)²

⇒ 9 = AC² + 16 + x² − 8x

⇒ AC² = − x² − 7 + 8x ........(ii)

From (i) and (ii), we get

25 – x² = − x² − 7 + 8x

8x = 32

x = 4

Hence, O'C = 4 - 4 = 0 cm i.e. 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

Also, AC² = 25 – x²

= 25 – 4²

= 25 − 16

= 9

⇒ AC = 3 cm

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