Question : 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:
Option 1: $\frac{4 \sqrt{13}}{3}$
Option 2: $\frac{2 \sqrt{14}}{3}$
Option 3: $\frac{2 \sqrt{13}}{3}$
Option 4: $\frac{4 \sqrt{14}}{3}$
Correct Answer: $\frac{4 \sqrt{14}}{3}$
Solution :
Given, AP = 5 cm, AQ = 3 cm and PQ = 6 cm
Let PM = x,
⇒ MQ = (6 – x)
In $\triangle$AMP
(PA)$^2$ = (AM)$^2$ + (PM)$^2$
⇒ 5
2
= (AM)$^2$ + x$^2$
⇒ (AM)$^2$ = 25 – x$^2$ -------------(1)
In $\triangle$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 equation (1) and equation (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 = $\sqrt{\frac{56}{9}}$ = $\frac{2\sqrt{14}}{3}$
As we know,
AB = 2AM = 2 ×$\frac{2\sqrt{14}}{3}$= $\frac{4\sqrt{14}}{3}$
Hence, the correct answer is $\frac{4\sqrt{14}}{3}$.
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