Question

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

• A. $\frac{\sqrt{3}}{2}$
• B. $\sqrt{3}$
• C. $2\sqrt{3}$
• D. $6$

Answer 1

The correct option is C $2\sqrt{3}$

$x^2+y^2=4$

Differentiating wrt x, we get

$\frac{dy}{dx}=-\frac{x}{y},m=\frac{-1}{\sqrt{3}}$

Equation of tangent: $y-\sqrt{3}=\frac{-1}{\sqrt{3}}(x-1)$

$\Rightarrow \sqrt{3}y-3=-x+1$

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

$Area=\left|\frac{y_1^2(1+m^2)}{2m}\right|=2\sqrt{3}$