If x^2+y^2-a^2+lambda(x cos alpha+y sin alpha | vedclass

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Question

(C) The equation of the family of circles passing through the intersection of the circle $x^2+y^2-a^2=0$ and the line $x \cos \alpha+y \sin \alpha-p=0

Vedclass · Accepted Answer

$-2p$

Vedclass · Answer

(C) The equation of the family of circles passing through the intersection of the circle $x^2+y^2-a^2=0$ and the line $x \cos \alpha+y \sin \alpha-p=0$ is given by $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$.For this to be the smallest circle,its center must lie on the line $x \cos \alpha+y \sin \alpha=p$.The center of the circle $x^2+y^2+\lambda x \cos \alpha+\lambda y \sin \alpha-(a^2+\lambda p)=0$ is $\left(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2}\right)$.Substituting the center into the line equation $x \cos \alpha+y \sin \alpha=p$:$\left(-\frac{\lambda \cos \alpha}{2}\right) \cos \alpha + \left(-\frac{\lambda \sin \alpha}{2}\right) \sin \alpha = p$$-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = p$$-\frac{\lambda}{2} (1) = p$$\lambda = -2p$.

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

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