Two circles of radii 10 cm and 17 cm intersecting each other at two points and the distance between their centres is 21 cm. Find the length of the common chord.

Let the length of OC be x cm and that of O'C be (21 - x)cm.

Since the perpendicular to a chord from the center of a circle bisects the chord, we use the Pythagorean theorem in two right triangles.

From the right triangle OAC, we have:

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 17² = x² + AC²

⇒ AC² = 289 - x² ……………….(1)

From the right triangle O'AC, we have:

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

⇒ 10² = (21 - x)² + AC²

⇒ 100 = 441 + x² - 42x + AC²

Using the equation (1):

⇒ 100 = 441 + x² - 42x + 289 - x²

⇒ 100 = 730 - 42x

⇒ 42x = 730 - 100

⇒ 42x = 630

⇒ x = $\dfrac{630}{42}$ = 15

Thus, OC = 15 cm and O'C - 21 - 15 = 6 cm.

Substituting the value of x in equation (1), we get:

⇒ AC² = 289 - x²

⇒ AC² = 289 - 15²

⇒ AC² = 289 - 225 = 64

⇒ AC = $\sqrt{64}$ = $\pm$ 8

Since the chord is bisected, length of common chord = 2 $\times$ AC = 2 $\times$ AC = 16 cm.

Hence, the length of the common chord = 16 cm.