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

यदि वृत्त \(x^2+y^2+6 x+8 y+16=0\) तथा \(x ^2+ y ^2+2(3-\sqrt{3}) x + x +2(4-\sqrt{6}) y\) \(= k +6 \sqrt{3}+8 \sqrt{6}, k > 0\) बिंदु \(P (\alpha, \beta)\) पर अंत: स्पर्श करते हैं, तो \((\alpha+\sqrt{3})^2+(\beta+\sqrt{6})^2\) बराबर है \(..............\)

  1. A \(24\)
  2. B \(298\)
  3. C \(25\)
  4. D \(56\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(25\)

Step-by-step Solution

Detailed explanation

The circle \(x^{2}+y^{2}+6 x+8 y+16=0\) has centre \((-3,-4)\) and radius 3 units. The circle \(x^{2}+y^{2}+2(3-\sqrt{3}) x+2(4-\sqrt{6}) y=\) \(k+6 \sqrt{3}+8 \sqrt{6}, k>0\) has centre \((\sqrt{3}-3, \sqrt{6}-4)\) and radius \(\sqrt{k+34}\) \(\because \quad\) These two circles…
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