Show that the line $\;$ $3x = y + 13$ $\;$ touches the circle $\;$ $x^2 + y^2 - 4x - 6y + 3 = 0$ $\;$ and find the point of contact.
Given line: $\;\;$ $3x = y + 13$ $\;\;\; \cdots \; (1a)$
i.e. $\;$ $3x - y - 13 = 0$ $\;\;\; \cdots \; (1b)$
Given circle: $\;\;$ $x^2 + y^2 - 4x - 6y + 3 = 0$ $\;\;\; \cdots \; (2)$
Comparing equation $(2)$ with the standard equation of a circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = -2, \; f = -3, \; c = 3$
$\therefore \;$ Center of given circle $= C = \left(-g, -f\right) = \left(2, 3\right)$
Radius of given circle $= r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 3} = \sqrt{10}$
Perpendicular distance from the center $C \left(2, 3\right)$ of the circle to line [equation $(1b)$] is
$= d = \left|\dfrac{3 \times 2 - 1 \times 3 - 13}{\sqrt{3^2 + \left(-1\right)^2}}\right|$
i.e. $\;$ $d = \left|\dfrac{6 - 3 - 13}{\sqrt{10}}\right| = \dfrac{10}{\sqrt{10}} = \sqrt{10}$
$\implies$ Perpendicular distance from the center $C$ of the circle to the given line $=$ radius of given circle
i.e. $\;$ The given line touches the circle
Now, from equation $(1a)$, $\;$ $y = 3x - 13$
Substituting the value of $y$ in equation $(2)$ gives
$x^2 + \left(3x - 13\right)^2 - 4x - 6 \left(3x - 13\right) + 3 = 0$
i.e. $\;$ $x^2 + 9x^2 - 78x + 169 - 4x - 18x + 78 + 3 = 0$
i.e. $\;$ $10 x^2 - 100 x + 250 = 0$
i.e. $\;$ $x^2 - 10 x + 25 = 0$
i.e. $\;$ $\left(x - 5\right)^2 = 0$
i.e. $\;$ $x = 5$
Substituting the value of $x$ in equation $(1a)$ gives
$y = 3x - 13 = 3 \times 5 - 13 = 2$
$\therefore \;$ The point of contact is $\;$ $\left(x, y\right) = \left(5, 2\right)$