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 $x\cos \alpha +y\sin \alpha =p,0<p<a$, then $\lambda =$

• 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
$\frac{-\lambda {\cos }^2\alpha }{2}-\frac{\lambda {\sin }^2\alpha }{2}=p$
$\Rightarrow \frac{-\lambda }{2}=p$
$\Rightarrow \lambda =-2p$