Find the angle between the tangents drawn from $\left(3, 4\right)$ to the circle $x^2 + y^2 - 4x - 2y - 4 = 0$.
External point: $\;$ $P \left(3, 4\right)$
Equation of circle: $\;$ $x^2 + y^2 - 4x - 2y - 4 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of a circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = -2, \; f = -1, \; c = -4$
$\therefore \;$ Center of circle $= C = \left(-g, -f\right) = \left(2, 1\right)$
Radius of circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 + 4} = 3$
Let $\theta$ be the angle between the two tangents drawn from the external point $P$.
Then, $\;\;$ $\sin \left(\dfrac{\theta}{2}\right) = \dfrac{r}{CP}$
i.e. $\;$ $\sin \left(\dfrac{\theta}{2}\right) = \dfrac{3}{\sqrt{\left(3 - 2\right)^2 + \left(4 - 1\right)^2}} = \dfrac{3}{\sqrt{1 + 9}} = \dfrac{3}{\sqrt{10}}$
i.e. $\;$ $\dfrac{\theta}{2} = \sin^{-1} \left(\dfrac{3}{\sqrt{10}}\right)$
i.e. $\;$ $\theta = 2 \sin^{-1} \left(\dfrac{3}{\sqrt{10}}\right)$