Find the tangents to the circle $x^2 + y^2 + 8x - 4y = 5$ parallel to the line $2x + 3y + 5 = 0$.
Equation of given circle: $\;\;\;$ $x^2 + y^2 + 8x - 4y = 5$
i.e. $\;$ $x^2 + y^2 + 8x - 4y - 5 = 0$ $\;\;\; \cdots \; (1)$
Comparing with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = 4, \; f = -2, \; c = -5$
Radius of the circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{16 + 4 + 5} = 5$
Equation of straight line: $\;\;\;$ $2x + 3y + 5 = 0$
Slope of the given straight line $= m = \dfrac{-2}{3}$
Any line parallel to the given straight line is $\;$ $2x + 3y + c_1 = 0$ $\;\;\; \cdots \; (2)$
Slope of line parallel to the given line $= m = \dfrac{-2}{3}$
Condition that the line $\;$ $y = mx + c_1$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ is
$c_1 = \pm r \sqrt{1 + m^2}$
i.e. $\;$ $c_1 = \pm 5 \sqrt{1 + \dfrac{4}{9}} = \pm \dfrac{5 \sqrt{13}}{3}$
$\therefore \;$ The equations of tangents are
$2x + 3y \pm \dfrac{5 \sqrt{13}}{3} = 0$
i.e. $\;$ $6x + 9y \pm 5 \sqrt{13} = 0$