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 $\frac{4\sqrt{14}}{3}$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : Two circles of radii 15 cm and 10 cm intersect each other and the length of their common chord is 16 cm. What is the distance (in cm) between their centres?
Question : What is the length (in cm) of the transverse common tangent between two circles with radii 6 cm and 4 cm, given that the distance between their centres is 14 cm?
Question : Two circles of radius 13 cm and 15 cm intersect each other at points A and B. If the length of the common chord is 12 cm, then what is the distance between their centres?
Question : Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Then the distance between their centres is:
Question : Two circles with their centres at O and P and radii 8 cm and 4 cm respectively touch each other externally. The length of their common tangent is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile