#### 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.

#### Solution

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 *x* − 4

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} + (*x* − 4)^{2}

⇒ 9 = AC^{2} + *x*^{2} + 16 − 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.

Length of the common chord AB = 2 O'A = (2 × 3) cm = 6 cm