If Δ 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 Δ = 

$2\sqrt{3}$

Equation of the tangent at $(1, \sqrt{3})$ is $x + \sqrt{3}y - 4 = 0$ …(i)

Equation of normal at $(1, \sqrt{3})$ is $\sqrt{3}x - y = 0$ …(ii)

Tangent cuts x-axis is y = 0 …(iii)

Point in the intersection of (i) and (ii) is (1, 3). Point of intersection of (i) and (iii) is (4, 0)

Point of intersection of (ii) and (iii) is (0, 0)

Area of the triangle = $\frac{1}{2}|0-4\sqrt{3}|=2\sqrt{3}$ sq units