Question

The area of the triangle formed by the positive x-axis and the normal and tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4$ at $\left(1,\sqrt{3}\right)$ is

• A. $\sqrt{3}$
• B. $2\sqrt{3}$
• C. $4\sqrt{3}$
• D. none of these

Answer 1

The correct option is B

$2\sqrt{3}$

Explanation for the correct option:

Given,

${\mathrm{x}}^2+{\mathrm{y}}^2=4$

Point $\left(1,\sqrt{3}\right)$

From the above figure,

$\mathrm{Area}=ar\left(\mathrm{OPA}\right)$

Tangent at

$\mathrm{P}(1,\sqrt{3})\Rightarrow \mathrm{y}-\sqrt{3}=-\frac{1}{\sqrt{3}}(\mathrm{x}-1)$

$\Rightarrow \mathrm{x}+\sqrt{3}\mathrm{y}=4$

When, $\mathrm{y}=0$ on tangent, $\mathrm{x}=4$

So, $\mathrm{A}=(4,0)$

Applying area of triangle,

$\mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$\mathrm{OA}=4;\mathrm{PM}=\sqrt{3}$

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\left(\mathrm{PM}\right)\times \left(\mathrm{OA}\right) \\ =\frac{1}{2}\times \sqrt{3}\times 4 \\ =2\sqrt{3} \end{gathered}$$

Hence, the correct answer is option (B).