The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^2+y^2=4$

at $(1,\sqrt{3})$ is….

The equation of tangent to the circle $x^2+y^2=4$ at $(1,\sqrt{3})$ is

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

The equation of normal to the circle $x^2+y^2=4$ at $(1,\sqrt{3})$ is

$y=\sqrt{3}x$

Thus, area of the triangle OPA

$\begin{array}{l}=\frac{1}{2}\times OA\times PM\\ =\frac{1}{2}\times 4\times \sqrt{3}\\ =2\sqrt{3}\text{ sq. u. }\end{array}$