Find the condition that the straight line $\;$ $x \cos \alpha + y \sin \alpha - p = 0$ $\;$ should touch the circle $\;$ $x^2 + y^2 - 2 \mu x = a^2$, $\;$ $\mu$ being a variable quantity. If $\mu_1$ and $\mu_2$ be the values of $\mu$, show that $\;$ $\mu_1 \mu_2 = \left(a^2 - p^2\right) \text{cosec}^2 \alpha$
Given circle: $\;$ $x^2 + y^2 - 2 \mu x = a^2$
i.e. $\;$ $x^2 + y^2 - 2 \mu x - a^2 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = - \mu, \; f = 0, \; c = - a^2$
$\therefore \;$ Center of the circle $= \left(-g, -f\right) = \left(\mu, 0\right)$
Radius of the circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{\mu^2 + a^2}$
Given line: $\;$ $x \cos \alpha + y \sin \alpha - p = 0$ $\;\;\; \cdots \; (2)$
Given line [equation $(2)$] is a tangent to the given circle [equation $(1)$] if
distance of the line from the center $=$ radius of the circle
i.e. $\;$ $\left|\dfrac{\mu \cos \alpha + 0 - p}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}\right| = \sqrt{\mu^2 + a^2}$
i.e. $\;$ $\mu^2 \cos^2 \alpha + p^2 - 2 p \mu \cos \alpha = \mu^2 + a^2$
i.e. $\;$ $\mu^2 \left(\cos^2 \alpha - 1\right) - 2 p \mu \cos \alpha + p^2 - a^2 = 0$
i.e. $\;$ $- \mu^2 \sin^2 \alpha - 2 p \mu \cos \alpha + p^2 - a^2 = 0$
i.e. $\;$ $\mu^2 + 2 p \mu \dfrac{\cos \alpha}{\sin^2 \alpha} + \dfrac{a^2 - p^2}{\sin^2 \alpha} = 0$
i.e. $\;$ $\mu^2 + 2 p \mu \cot \alpha \text{ cosec} \alpha + \left(a^2 - p^2\right) \text{cosec}^2 \alpha = 0$ $\;\;\; \cdots \; (3)$
Equation $(3)$ is the required condition to be satisfied so that line $(2)$ is a tangent to the circle $(1)$.
Equation $(3)$ is a quadratic equation in $\mu$.
If $\mu_1, \; \mu_2$ are the values of $\mu$, then
Product of roots of equation $(3)$ $= \mu_1 \mu_2 = \dfrac{\left(a^2 - p^2\right) \text{cosec}^2 \alpha}{1} = \left(a^2 - p^2\right) \text{cosec}^2 \alpha$