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  • The common chord of  <img alt="\mathrm{x^2+y^2-4 x-4 y=0 \text { and } x^2+y^2=16}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-4%20x-4%20y%3D0%2

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

Option: 1

\mathrm{\pi / 6}


Option: 2

\mathrm{\pi / 4}


Option: 3

\mathrm{\pi / 3}


Option: 4

\mathrm{\pi / 2}


Answers (1)

best_answer

The equation of the commo chord of the circles  \mathrm{x^2+y^2-4 x-4 y=0 \text { and } x^2+y^2=16 \text { is } x+y=4} which

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

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

Hence (d) is the correct answer.

Posted by

Rishi

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