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TS EAMCET · Maths · Circle

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

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

Answer & Solution

Correct Answer

(C) \(-2 p\)

Step-by-step Solution

Detailed explanation

Equation of circle \(x^2+y^2-a^2+\lambda\) \((x \cos \alpha+y \sin \alpha-p)=0\) is the smallest circle, then centre \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) lies on the line \(x \cos \alpha+y \sin \alpha=p\) \(\ldots(\mathrm{i})\)…