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AP EAMCET · Maths · Circle

If the smallest circle through the points of intersection of \(x^2+y^2=a^2\) and \(x \cos \alpha+y \sin \alpha=p, 0 < p < a\) is \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) then \(\lambda=\)

  1. A \(1\)
  2. B \(-1\)
  3. C \(-\mathrm{p}\)
  4. D \(-2 p\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-2 p\)

Step-by-step Solution

Detailed explanation

\(x^2+y^2 + \lambda x \cos \alpha + \lambda y \sin \alpha - (a^2+\lambda p) = 0\) Center: \((-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2})\) For smallest circle, center lies on chord:…