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

• A

  $\frac{1}{3}$
• Correct answer
  B

  $\frac{2}{\sqrt{3}}$
• C

  $\frac{4}{\sqrt{3}}$
• D

  $\frac{1}{\sqrt{3}}$

Solution

Equation tangent at $P\left(\sqrt{3},1\right)$
$x\cdot \sqrt{3}+y\cdot 1=4$
x-intercept of this tangent is $\frac{4}{\sqrt{3}}$.
Hence, area of $\Delta =\frac{1}{2}\cdot \frac{4}{\sqrt{3}}\cdot 1=\frac{2}{\sqrt{3}}$

Hence, the correct answer is option: 'B'.