Find the equation to the circle which has its center at $\left(1, -2\right)$ and touches the line $7x - 24y - 5 = 0$.
Center of the required circle $= \left(h, k\right) = \left(1, -2\right)$
The required circle touches the line $\;$ $7x - 24y -5 = 0$
$\therefore \;$ Perpendicular distance from the center $\left(0, 0\right)$ to the point where given line touches the circle is the radius of the required circle.
i.e. $\;$ $r = \left|\dfrac{7 \times 1 - 24 \times \left(-2\right) - 5}{\sqrt{7^2 + \left(-24\right)^2}}\right|$
i.e. $\;$ $r = \left|\dfrac{7 + 48 - 5}{\sqrt{49 + 576}}\right| = \left|\dfrac{50}{\sqrt{625}}\right|$
i.e. $\;$ $r = \dfrac{50}{25} = 2$
Equation of circle with center at $\left(h, k\right)$ and radius $= r$ is $\;$ $\left(x - h\right)^2 + \left(y - k\right)^2 = r^2$
$\therefore \;$ Equation of required circle is
$\left(x - 1\right)^2 + \left(y - 2\right)^2 = 2^2$
i.e. $\;$ $x^2 + y^2 - 2x + 4y + 1 = 0$