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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

વર્તુળ \((x-\alpha)^2+(y-\beta)^2=50\) જ્યાં \(\alpha, \beta>0\) ધ્યાને લો. જો વર્તુળ, એ રેખા \(y+x=0\) ને બિંદુ \(P\) આગળ સ્પર્શે, જેનું ઊગમબિંદુ થી અંતર \(4 \sqrt{2}\) છે, તો \((\alpha+\beta)^2 =\) ...........

  1. A \(103\)
  2. B \(102\)
  3. C \(55\)
  4. D \(100\)
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Correct Answer

(D) \(100\)

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Detailed explanation

\( S:(x-\alpha)^2+(y-\beta)^2=50 \) \( C P=r \) \( \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} \) \( \Rightarrow(\alpha+\beta)^2=100\)
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