Find the tangents to the circle $x^2 + y^2 = 81$ parallel to the line $3x + 4y = 13$.
Equation of given circle: $\;\;\;$ $x^2 + y^2 = 81$ $\;\;\; \cdots \; (1)$
Comparing with the standard equation of circle $\;$ $x^2 + y^2 = r^2$ $\;$ gives
$r^2 = 81$
Equation of straight line: $\;\;\;$ $3x + 4y = 13$
Any line parallel to the given straight line is $\;$ $3x + 4y + c = 0$ $\;\;\; \cdots \; (2)$
Equation $(2)$ will touch the given circle, whose center is $= \left(0, 0\right)$ and radius is $= 9$ if
$\left|\dfrac{3 \times 0 + 4 \times 0 + c}{\sqrt{3^2 + 4^2}}\right| = 9$
i.e. $\;$ $\dfrac{\pm c}{5} = 9$
i.e. $\;$ $c = \pm 45$
$\therefore \;$ The equations of tangents are
$3x + 4y + 45 = 0$ $\;\;$ and $\;\;$ $3x + 4y - 45 = 0$
OR
Condition that the line $\;$ $y = mx + c$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ is
$c = \pm r \sqrt{1 + m^2}$
i.e. $\;$ $c = \pm 9 \sqrt{1 + \left(\dfrac{-3}{4}\right)^2}$
i.e. $\;$ $c = \pm 9 \sqrt{\dfrac{25}{16}}$
i.e. $\;$ $c = \pm \dfrac{45}{4}$
$\therefore \;$ The equations of tangents are
$y = \dfrac{-3}{4}x \pm \dfrac{45}{4}$
i.e. $\;$ $3x + 4y \pm 45 = 0$