Find the condition that the straight line $x \cos \alpha + y \sin \alpha = p$ may touch the circle $x^2 + y^2 = r^2$.
Equation of given circle: $\;\;\;$ $x^2 + y^2 = r^2$ $\;\;\; \cdots \; (1)$
Equation of straight line: $\;\;\;$ $x \cos \alpha + y \sin \alpha = p$ $\;\;\; \cdots \; (2)$
i.e. $\;$ $y = \dfrac{p - x \cos \alpha}{\sin \alpha}$
The $x$ coordinates of the points of intersection of equations $(1)$ and $(2)$ are given by the equation
$x^2 + \left(\dfrac{p - x \cos \alpha}{\sin \alpha}\right)^2 = r^2$
i.e. $\;$ $x^2 \sin^2 \alpha + p^2 - 2px \cos \alpha + x^2 \cos^2 \alpha = r^2 \sin^2 \alpha$
i.e. $\;$ $x^2 - 2p x \cos \alpha + p^2 - r^2 \sin^2 \alpha = 0$ $\;\;\; \cdots \; (3)$
Equation $(2)$ will be a tangent to $(1)$ if the quadratic given by $(3)$ has equal roots.
i.e. $\;$ if the discriminant of quadratic equation $(3)$ is zero.
i.e. if $\;$ $4 p^2 \cos^2 \alpha - 4 \left(p^2 - r^2 \sin^2 \alpha\right) = 0 $
i.e. $\;$ $p^2 \left(1 - \cos^2 \alpha\right) = r^2 \sin^2 \alpha$
i.e. $\;$ $p^2 \sin^2 \alpha = r^2 \sin^2 \alpha$
i.e. $\;$ $p = \pm r$