• # question_answer Let a circle of radius $\frac{2}{\sqrt{3}}$ is touching the lines ${{x}^{2}}-4xy+{{y}^{2}}=0$in the first quadrant at points A and B. If area of triangle OAB (0 being the origin) is $\Delta$ then ${{\Delta }^{2}}$ equals A)  1                     B)  3 C)  2                                 D)  1

Here $\tan 2\theta \,=\frac{2\sqrt{4}-1}{2}\,=\sqrt{3}\,\Rightarrow \,\theta \,=\frac{\pi }{6}$ $\therefore \,\,$ Area of $\Delta OAB=\frac{1}{2}{{(r\cos \theta )}^{2}}(\sin 2\theta )$ $=\frac{1}{2}\,{{\left( r\sqrt{3} \right)}^{2}}\frac{\sqrt{3}}{2}\,=\frac{3\sqrt{3}{{r}^{2}}}{4}$ $={{\Delta }^{2}}=3\,\,\left( As\,r=\frac{2}{\sqrt{3}} \right)$