Two circles of radii 17 cm and 25 cm intersect each other at two points A and B. If the length of common chord AB of the circles is 30 cm, find the distance between the centres of the circles.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AC = CB = $\dfrac{30}{2}$ = 15 cm

From figure,

In right triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 25² = OC² + 15²

⇒ 625 = OC² + 225

⇒ OC² = 400

⇒ OC = $\sqrt{400}$ = 20 cm.

In right triangle O'AC,

⇒ O'A² = O'C² + AC² (By pythagoras theorem)

⇒ 17² = O'C² + 15²

⇒ 289 = O'C² + 225

⇒ O'C² = 64

⇒ O'C = $\sqrt{64}$ = 8 cm.

Distance between centers = OO' = OC + O'C = 20 + 8 = 28 cm.

Hence, distance between their centres = 28 cm.