The diameter and a chord of a 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 centre of the circle?
AB is the diameter and AC is the chord.
Draw OL ⊥ AC
Since OL ⊥ AC and hence it bisects AC, O is the centre of the circle.
Therefore, OA = 10 cm and AL = 6 cm
Now, in Rt. ΔOLA,
`AO^2 = AL^2 + OL^2`
`⇒ 10 ^2 = 6^2 + OL^2`
`⇒ OL^2 = 100 - 36 = 64`
⇒ OL = 8 cm
Therefore, chord is at a distance of 8 cm from the centre of the circle.