JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If the circles \(x^{2}+y^{2}+6 x+8 y+16=0\) and \(x^{2}+y^{2}+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y\) \(= k +6 \sqrt{3}+8 \sqrt{6}, k >0\), touch internally at the point \(P(\alpha, \beta)\), then \((\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}\) is equal to \(\dots\dots\)
- A \(24\)
- B \(298\)
- C \(25\)
- D \(56\)
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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