CBSE Class 12 · 2015 · Area Under CurvesLong Answer

The tangent to $x^2+y^2=4$ at $(1,\sqrt{3})$ is $x(1)+y(\sqrt{3})=4$, i.e. $x+\sqrt{3}y=4$; it cuts the $x$-axis at $(4,0)$.
The normal passes through the centre $O(0,0)$ and $(1,\sqrt{3})$, i.e. $y=\sqrt{3}x$; it meets the $x$-axis at the origin.
So the triangle is formed by the origin $(0,0)$, the point $(4,0)$ on the $x$-axis, and the point of contact $(1,\sqrt{3})$. Taking the base on the $x$-axis from $0$ to $4$ and height $=\sqrt{3}$:

equivalently $\frac{1}{2}\cdot 4\cdot \sqrt{3}=2\sqrt{3}$. Hence the area is $2\sqrt{3}$ square units.