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(D) The equation of any circle passing through the intersection of the circle $S: x^2+y^2-a^2=0$ and the line $L: x \cos \alpha+y \sin \alpha-p=0$ is

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$-2p$

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(D) The equation of any circle passing through the intersection of the circle $S: x^2+y^2-a^2=0$ and the line $L: x \cos \alpha+y \sin \alpha-p=0$ is given by $S+\lambda L=0$,which is $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$.This can be rewritten as $x^2+\lambda x \cos \alpha+y^2+\lambda y \sin \alpha-(a^2+\lambda p)=0$.The center of this circle is $(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2})$.The smallest circle passing through the intersection points of a circle and a line is the circle that has the chord of intersection as its diameter.The line $x \cos \alpha+y \sin \alpha=p$ is the chord of intersection.The center of the circle must lie on this line.Substituting the center into the line equation: $(-\frac{\lambda \cos \alpha}{2}) \cos \alpha + (-\frac{\lambda \sin \alpha}{2}) \sin \alpha = p$.$-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = p$.$-\frac{\lambda}{2} = p$.$\lambda = -2p$.

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

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