The length of the common chord of two circles of radii 15 cm and 13 cm, whose centres are 14 cm apart, is:
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Correct answer: D
Explanation
The correct answer is option 4 i.e. 24 cm.
PO = 14 cm, r1 = 13 cm, r2 = 15 cm
Let's assume OC = x, PC = 14 - x
in triangle ACP
AP2 = PC2 + AC2
⇒ 169 = (14 - x)2 + AC2
⇒ AC2 = 169 - 196 + 28x - x2
⇒ AC2 = -27 + 28x - x2...............(1)
in triangle ACO
AO2 = OC2 + AC2
⇒ 225 = x2 + AC2
⇒ AC2 = 225 - x2..........................(2)
From eq 1 and 2
⇒ -27 + 28x - x2 = 225 - x2
⇒ 28x = 252
x = 252/28 = 9
So, AC2 = 225 - 92 = 144
AC = 12
The length of the common chord = 2 × 12 = 24