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 \(\lambda=\)
- A 1
- B \(-p\)
- C \(-2 p\)
- D \(-3 p\)
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 \(x \cos \alpha+y \sin \alpha=p\) \(\ldots(\mathrm{i})\)…
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