*two circle of radii 5cm and 3cm intersect at two points and the distance between their centres is 4 cm .find the length of common chord.*

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,

OA^{2 }= AC^{2} + OC^{2}

⇒ 5^{2} = AC^{2} +* x*^{2}

⇒ 25 − *x*^{2} = AC^{2} ... (1)

In ΔO'AC,

O'A^{2} = AC^{2} + O'C^{2}

⇒ 3^{2} = AC^{2} + (4 − *x*)^{2}

⇒ 9 = AC^{2} + 16 + *x*^{2} − 8*x*

⇒ AC^{2} = − *x*^{2} − 7 + 8*x* ... (2)

From equations (1) and (2), we obtain

25 − *x*^{2 }= − *x*^{2} − 7 + 8*x*

8*x* = 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.

AC^{2} = 25 − *x*^{2} = 25 − 4^{2} = 25 − 16 = 9

∴ AC = 3 m

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