Question

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

• A. 1
• B. -p
• C. -2p
• D. -3p

Answer 1

Answer: (C)

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
$x \cos \alpha+y \sin \alpha=p$....(i)
$\therefore $ Put centre in the line (i), then we getting
$\frac{-\lambda \cos ^{2} \alpha}{2}-\frac{\lambda \sin ^{2} \alpha}{2}=p$
$\Rightarrow \frac{-\lambda}{2}=p$
$\Rightarrow \lambda=-2 p$