Show that the straight line $3x - 4y + 10 = 0$ touches the circle $x^2 + y^2 = 4$ and find the point of contact.
Given line: $\;\;$ $3x - 4y + 10 = 0$
i.e. $\;$ $y = \dfrac{3}{4} x + \dfrac{10}{4}$
i.e. $\;$ $y = \dfrac{3}{4} x + \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of line $\;$ $y = mx + c$ $\;$ gives
$c = \dfrac{5}{2}, \;\; m = \dfrac{3}{4}$
Given circle: $\;\;$ $x^2 + y^2 = 4$ $\;\;\; \cdots \; (2)$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 = r^2$ $\;$ gives
$r^2 = 4$
Condition that the line $\;$ $y = mx + c$ $\;$ is a tangent to the circle $\;$ $x^2 + y^2 = r^2$ $\;$ is $\;\;$ $c^2 = r^2 \left(1 + m^2\right)$
Now,
$c^2 = \left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$ $\;\;\; \cdots \; (3a)$
$r^2 \left(1 + m^2\right) = 4 \left(1 + \dfrac{9}{16}\right) = 4 \times \dfrac{25}{16} = \dfrac{25}{4}$ $\;\;\; \cdots \; (3b)$
$\therefore \;$ From equations $(3a)$ and $(3b)$, $\;\;$ $c^2 = r^2 \left(1 + m^2\right)$
$\implies$ The given line is a tangent to the given circle.
In view of equation $(1)$ equation $(2)$ becomes
$x^2 + \left(\dfrac{3}{4}x + \dfrac{5}{2}\right)^2 = 4$
i.e. $\;$ $x^2 + \dfrac{9}{16} x^2 + \dfrac{25}{4} + 2 \times \dfrac{3x}{4} \times \dfrac{5}{2} = 4$
i.e. $\;$ $\dfrac{25}{16} x^2 + \dfrac{15}{4} x + \dfrac{9}{4} = 0$
i.e. $\;$ $25 x^2 + 60 x + 36 = 0$
i.e. $\;$ $\left(5x + 6\right)^2 = 0$
$\implies$ $x = \dfrac{-6}{5}$
Substituting the value of $x$ in equation $(1)$ gives
$y = \dfrac{3}{4} \times \left(\dfrac{-6}{5}\right) + \dfrac{5}{2} = \dfrac{8}{5}$
$\therefore \;$ The point of contact is $\;\;$ $\left(\dfrac{-6}{5}, \dfrac{8}{5}\right)$