The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the center of the circle?

In the figure below, AB is the diameter and AC as the chord.

Now, draw OL ⊥ AC

Since, O is the centre of the circle and OL ⊥ AC.

∴ L bisects AC.

∴ AL = 6 cm and OA = radius = 10 cm.

Now, in right ∆OLA

⇒ OA² = AL² + OL² [By Pythagoras Theorem]

⇒ 10² = 6² + OL²

⇒ OL² = 100 - 36

⇒ OL² = 64

⇒ OL = $\sqrt{64}$ = 8 cm.

Hence, the chord is at a distance of 8 cm from the centre of the circle.