Question

The tangent and the normal lines at the point $(\sqrt{3},1)$ to the circle $x^2+y^2=4$ and the $x$-axis forms a triangle. The area of this triangle (in square units) is :

• A. $\frac{4}{\sqrt{3}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{2}{\sqrt{3}}$


Equation of the tangent at $A(\sqrt{3},1)$ is
$\sqrt{3}x+y=4\Rightarrow \frac{x}{4/\sqrt{3}}+\frac{y}{4}=1$
Therefore, the coordinates of point $B$
$=(\frac{4}{\sqrt{3}},0)$

$\therefore$ Area of $△OAB$
$=\frac{1}{2}\times OB\times h=\frac{1}{2}\times \frac{4}{\sqrt{3}}\times 1=\frac{2}{\sqrt{3}}$