For what possible value of p is the line x co | vedclass

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(C) The given equation of the circle is $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2

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$0$ and $2q$

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(C) The given equation of the circle is $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -q \cos \alpha$ and $f = -q \sin \alpha$. The center of the circle is $(-g, -f) = (q \cos \alpha, q \sin \alpha)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{q^2 \cos^2 \alpha + q^2 \sin^2 \alpha} = \sqrt{q^2(1)} = |q|$. For the line $x \cos \alpha + y \sin \alpha - p = 0$ to be a tangent to the circle,the perpendicular distance from the center to the line must be equal to the radius $r$. Distance $d = \frac{|(q \cos \alpha)(\cos \alpha) + (q \sin \alpha)(\sin \alpha) - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |q(\cos^2 \alpha + \sin^2 \alpha) - p| = |q - p|$. Setting $d = r$,we get $|q - p| = |q|$. This implies $q - p = q$ or $q - p = -q$. Therefore,$p = 0$ or $p = 2q$.

For what possible value of $p$ is the line $x \cos \alpha + y \sin \alpha = p$ a tangent to the circle $x^2 + y^2 - 2qx \cos \alpha - 2qy \sin \alpha = 0$?

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