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(D) The equation of a circle passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$. Here,$S = x^2 +

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$x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$

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(D) The equation of a circle passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$. Here,$S = x^2 + y^2 - a^2 = 0$ and $L = x \cos \alpha + y \sin \alpha - p = 0$. So,the equation of the circle is $(x^2 + y^2 - a^2) + \lambda(x \cos \alpha + y \sin \alpha - p) = 0$. Since $AB$ is the diameter,the center of this circle must lie on the line $L = 0$. The center of the circle $(x^2 + y^2 + \lambda x \cos \alpha + \lambda y \sin \alpha - a^2 - \lambda p) = 0$ is $(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2})$. Substituting this into $x \cos \alpha + y \sin \alpha - p = 0$: $(-\frac{\lambda \cos \alpha}{2}) \cos \alpha + (-\frac{\lambda \sin \alpha}{2}) \sin \alpha - p = 0$ $-\frac{\lambda}{2}(\cos^2 \alpha + \sin^2 \alpha) = p$ $-\frac{\lambda}{2} = p \Rightarrow \lambda = -2p$. Substituting $\lambda = -2p$ into the equation: $x^2 + y^2 - a^2 - 2p(x \cos \alpha + y \sin \alpha - p) = 0$.

The straight line $x \cos \alpha + y \sin \alpha = p$ cuts the circle $x^2 + y^2 - a^2 = 0$ at $A$ and $B$. Then the equation of the circle having $AB$ as diameter is

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