A circle with center at the origin, is described so as to touch the straight line $2y = x + 6$. Find its radius.
Let the equation of the required circle with center at the origin, i.e. $\;$ $\left(0, 0\right)$ $\;$ and radius $\;$ $r$ $\;$ be
$x^2 + y^2 = r^2$ $\;\;\; \cdots \; (1)$
Given line: $\;$ $2y = x + 6$
i.e. $\;$ $y = \dfrac{1}{2} x + 3$ $\;\;\; \cdots \; (2)$
Comparing equation $(2)$ with the standard equation of a straight line $\;$ $y = mx + c$ $\;$ gives
Slope $ = m = \dfrac{1}{2}$, $\;$ intercept on the $Y$ axis $= c = 3$
Equation $(2)$ will be a tangent to the circle given by equation $(1)$ if
$c^2 = \left(1 + m^2\right) r^2$
$\implies$ $r^2 = \dfrac{c^2}{1 + m^2}$
i.e. $\;$ $r^2 = \dfrac{3^2}{1 + \left(\dfrac{1}{2}\right)^2} = \dfrac{36}{5}$
$\therefore \;$ Radius of the circle $= r = \dfrac{6}{\sqrt{5}}$