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The common chord of \mathrm{x^2+y^2-4 x-4 y=0} and \mathrm{x^2+y^2=16} subtends at the origin an angle equal to

Option: 1

\mathrm{\frac{\pi}{6}}


Option: 2

\mathrm{\frac{\pi}{4}}


Option: 3

\mathrm{\frac{\pi}{3}}


Option: 4

\mathrm{\frac{\pi}{2}}


Answers (1)

best_answer

The equation of the common chord of the circles 

\mathrm{x^2+y^2-4 x-4 y=0 \text { and } x^2+y^2=16}  is x + y = 4 which meets the circle

\mathrm{x^2+y^2=16}  at points A(4, 0) and B(0, 4). Obviously \mathrm{\mathrm{OA} \perp \mathrm{OB}}. Hence the

common chord AB makes a right angle at the centre of the circle \mathrm{x^2+y^2=16}
 

Posted by

Irshad Anwar

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