Answer :

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

Also distance between their centers is 4 cm.

So OO^{'} = 4 cm

The diagram is shown below:

If two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of the common cord.

OO' will be the perpendicular bisector of chord AB

⇒ AC = CB

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} ..........(i)

Now, In ΔO'AC,

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

⇒ 3^{2} = AC^{2} + (4 − x)^{2}

⇒ 9 = AC^{2} + 16 + x^{2} − 8x

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

From (i) and (ii), we get

25 – x^{2} = − x^{2} − 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, AC^{2} = 25 – x^{2}

= 25 – 4^{2}

= 25 − 16

= 9

⇒ AC = 3 cm

⇒ Length of the common chord AB = 2 AC

= (2 × 3) cm

= 6 cm

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