Explanation

(c)
Since, $CA$ is perpendicular to the tangent
[image:11_image74.emf]
$\therefore\ \ \ r = \frac{|h - 0|}{\sqrt{1^{2} + 1^{2}}}$
$\Rightarrow \ \ \ h^{2} = 2r$ …(i)
Since, $CN$ is perpendicular to the chord of line,
$y = \frac{x}{\sqrt{3}}$
$\therefore\ CN = \frac{\left| \frac{h}{\sqrt{3}} - 0 \right|}{\sqrt{\frac{1}{3} + 1}} = \frac{h}{2}$
In $\mathrm{\Delta}BNC$, $r^{2} = 1^{2} + \left( \frac{h}{2} \right)^{2}$
$r^{2} = 1 + \frac{h^{2}}{4}$ …(ii)
From Eqs. (i) and (ii), we get
$r^{2} = 1 + \frac{2r^{2}}{4}$
$\Rightarrow \ \ \ \ r = \sqrt{2}$