Find the equation to the circle which has its center at $\left(0, 0\right)$ and touches the line $12x + 5y - 11 = 0$.
Center of the required circle $= \left(0, 0\right)$
The required circle touches the line $\;$ $12x + 5y - 11 = 0$
$\therefore \;$ 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. $\;$ $r = \left|\dfrac{12 \times 0 + 5 \times 0 - 11}{\sqrt{12^2 + 5^2}}\right|$
i.e. $\;$ $r = \dfrac{11}{\sqrt{144 + 25}} = \dfrac{11}{13}$
Equation of circle with center at $\left(0, 0\right)$ and radius $= r$ is $\;$ $x^2 + y^2 = r^2$
$\therefore \;$ Equation of required circle is
$x^2 + y^2 = \left(\dfrac{11}{13}\right)^2$
i.e. $\;$ $169 x^2 + 169 y^2 = 121$