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Question:

There is a common chord of 2 circles with radius 15 and 20. The distance between the two centres is 25. The length of the chord is

Explanation:

In ∆ABC,

AB² + AC² = BC²

∴∠BAC = 90°

Let BP = x

∴ PC = 25 − x

202 = x² + AP²                                        ...(i)

In ∆APC,

225 = AP² + (25 – x)²

225 = AP² + 625 + x² – 50x                   ...(ii)

Equating (i) and (ii), we get,

225 = 625 + 400 – 50x

∴ 50x = 800

∴ x = 16

AP² = 20² – 16² = 12²

∴ AP = 12

∴ Length of the chord = 2 × AP = 24

Hence, option (b).