Two circles of radii 5 cm and 3 cm intersect each other at A and B,and the distance between their centres is 6 cm. The length (in cm) of the common chord AB is:
Answer & explanation
Correct answer: option 3
Given, AP = 5 cm, AQ = 3 cm and PQ = 6 cm
Let PM = x,
= MQ = (6 - x)
In \(\Delta \)AMP
\( {(PA) }^{2 } \) = \( {(AM) }^{2 } \) + \( {(PM) }^{2 } \)
= \( {5 }^{2 } \) = \( {(AM) }^{2 } \) + \( {(X }^{2 } \)
= \( {(AM) }^{2 } \) = 25 - \( {X }^{2 } \) ..(1)
In \(\Delta \)AQM,
\( {(AQ) }^{2 } \) = \( {(AM) }^{2 } \) + \( {(6\;-\;X) }^{2 } \)
= \( {3 }^{2 } \) = \( {(AM) }^{2 } \) + 36 + \( {X }^{2 } \) - 12X
= \( {(AM) }^{2 } \) = 9 - 36 - \( {X }^{2 } \) + 12X ..(2)
From eq(1) and eq(2)
25 - \( {X }^{2 } \) = 9 - 36 - \( {X }^{2 } \) + 12X
12X = 25 + 27
12X = 52
X = \(\frac{52}{12}\)
X = \(\frac{13}{3}\)
From equation (1)
\( {(AM) }^{2 } \) = 25 - \( {(\frac{13}{3}) }^{2 } \) = 25 - \(\frac{169}{9}\) = \(\frac{225\;-\;169}{9}\) = \(\frac{56}{9}\)
AM = \(\frac{\sqrt {56 }}{9}\) = \(\frac{2\sqrt {14 }}{3}\)
As we know,
AB = 2AM = 2 x [\(\frac{2\sqrt {14 }}{3}\)] = \(\frac{4\sqrt {14 }}{3}\)
Therefore, AB is \(\frac{4\sqrt {14 }}{3}\).