The line x cos alpha + y sin alpha = p will b | vedclass

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(D) The line $x \cos \alpha + y \sin \alpha - p = 0$ is a tangent to the circle if the perpendicular distance from the center to the line is equal to

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$0$ or $2a$

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(D) The line $x \cos \alpha + y \sin \alpha - p = 0$ is a tangent to the circle if the perpendicular distance from the center to the line is equal to the radius of the circle.The given circle equation is $x^2 + y^2 - 2ax \cos \alpha - 2ay \sin \alpha = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get center $(g, f) = (a \cos \alpha, a \sin \alpha)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{a^2 \cos^2 \alpha + a^2 \sin^2 \alpha - 0} = a$.The perpendicular distance from $(a \cos \alpha, a \sin \alpha)$ to the line $x \cos \alpha + y \sin \alpha - p = 0$ is given by:$d = \frac{|(a \cos \alpha) \cos \alpha + (a \sin \alpha) \sin \alpha - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}$$d = |a(\cos^2 \alpha + \sin^2 \alpha) - p| = |a - p|$.Setting $d = r$,we get $|a - p| = a$.This implies $a - p = a$ or $a - p = -a$.Therefore,$p = 0$ or $p = 2a$.

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the circle $x^2 + y^2 - 2ax \cos \alpha - 2ay \sin \alpha = 0$,if $p = $

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