Find the value of $p$ so that the line $3x + 4y - p = 0$ is a tangent to $x^2 + y^2 - 64 = 0$.
Given equation of circle: $\;\;\;$ $x^2 + y^2 - 64 = 0$ $\;$ i.e. $\;$ $x^2 + y^2 = 64$ $\;\;\; \cdots \; (1)$
Comparing with the standard equation of circle: $\;\;\;$ $x^2 + y^2 = a^2$
gives $\;\;\;$ $a^2 = 64$ $\;\;\; \cdots \; (2)$
Equation of given line: $\;\;\;$ $3x + 4y - p = 0$ $\;$ i.e. $\;$ $y = \dfrac{-3}{4}x + \dfrac{p}{4}$ $\;\;\; \cdots \; (3)$
Comparing equation $(3)$ with the standard equation of line $\;$ $y = mx + c$ $\;$ gives
$m = \dfrac{-3}{4}$ $\;\;\; \cdots \; (4a)$; $\;$ $c = \dfrac{p}{4}$ $\;\;\; \cdots \; (4b)$
Condition for a given line to be a tangent to a circle is: $\;$ $c^2 = a^2 \left(1 + m^2\right)$ $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ In view of equations $(2)$, $(4a)$ and $(4b)$, equation $(5)$ becomes
$\dfrac{p^2}{16} = 64 \left[1 + \left(\dfrac{-3}{4}\right)^2\right]$
i.e. $\dfrac{p^2}{16} = 64 \left(1 + \dfrac{9}{16}\right)$
i.e. $\dfrac{p^2}{16} = 64 \times \dfrac{25}{16}$ $\implies$ $p = \pm 40$