Discussion

Explanation:

We can draw the following diagram from the infomation given.

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The perpendicular drawn on a chord from the center of the circle bisects the chord
⇒ BX = 16/2 = 8 and
⇒ DY = 12/2 = 6 and

Now, in ∆BOX
⇒ BO2 = BX2 + OX2
⇒ 102 = 82 + OX2
⇒ OX = 6

Also, in ∆BOX
⇒ DO2 = DY2 + OY2
⇒ 102 = 62 + OY2
⇒ OY = 8

XY = OX + OY = 6 + 8 = 14

Hence, option (c).

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