JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
A circle \(\mathrm{C}\) touches the line \(\mathrm{x}=2 \mathrm{y}\) at the point \((2,1)\) and intersects the circle \(C_{1}: x^{2}+y^{2}+2 y-5=0\) at two points \(\mathrm{P}\) and \(\mathrm{Q}\) such that \(\mathrm{PQ}\) is a diameter of \(\mathrm{C}_{1}\). Then the diameter of \(\mathrm{C}\) is :
- A \(7 \sqrt{5}\)
- B \(15\)
- C \(\sqrt{285}\)
- D \(4 \sqrt{15}\)
Answer & Solution
Correct Answer
(A) \(7 \sqrt{5}\)
Step-by-step Solution
Detailed explanation
\((\mathrm{x}-2)^{2}+(\mathrm{y}-1)^{2}+\lambda(\mathrm{x}-2 \mathrm{y})=0\) \(\mathrm{C}: \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}(\lambda-4)+\mathrm{y}(-2-2 \lambda)+5=0\) \(\mathrm{C}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{y}-5=0\) \(\mathrm{~S}_{1}-\mathrm{S}_{2}=0\)…
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