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Question
A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the center.
Solution
Let AB be the chord of length 24 cm and O be the center of the circle.
Let OC be the perpendicular drawn from O to AB.
We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.
∴ AC = CB = 12 cm
In OCA,
OA2 = OC2 + AC2 ....( By Pythagoras theorem )
=(5)2 + ( 12 )2 = 169
⇒ OA = 13 cm
∴ radius of the circle = 13 cm.
Let A ' B ' be the new chord at a distance of 12 cm from the center.
∴ ( OA' )2 = ( OC' )2 + ( A'C' )2
⇒ ( A'C' )2 = ( 13 )2 - ( 12 )2 = 25
∴ A'C' = 5 cm
Hence, length of the new chord = 2 x 5 = 10 cm.
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