Two circles of radii 5cm and …

Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'.

Hence OA = OB = 5 cm
➾ O'A = O'B
➾ 3 cm

OO' is the perpendicular bisector of chord AB.

Therefore, AC = BC

Given, OO' = 4 cm

Let OC = x

 Hence O'C = 4 − x

In right angled ΔOAC, by Pythagoras theorem OA²

➾ OC² + AC²

➾ 5² = x² + AC²

➾ AC² = 25 − x²à

(1) In right angled ΔO'AC, by Pythagoras theorem O'A²

➾ AC² + O'C²

➾ 3² = AC² + (4 – x)²

➾ 9 = AC² + 16 + x² − 8x

➾ AC² = 8x − x² − 7 à

 (2) From (1) and (2),

 we get 25 − x² = 8x − x² − 7 8x = 32

 Therefore, x = 4

 Hence, the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.

AC² = 25 − x²

➾ 25 − 4²

➾ 25 − 16 = 9

 Therefore, AC = 3 m Length of the common chord,

➾ AB = 2AC
➾ 6 cm