Question:
The tangent and the normal lines at the point $(\sqrt{3}, 1)$ to the circle $x^{2}+y^{2}=4$ and the $x$-axis form a triangle. The area of this triangle (in square units) is :
Correct Option: , 4
Solution:
Given $x^{2}+y^{2}=4$
equation of tangent
$\Rightarrow \sqrt{3} x+y=4$ ..(1)
Equation of normal
$x-\sqrt{3} y=0$ ....(2)
Coordinate of $\mathrm{T}\left(\frac{4}{\sqrt{3}}, 0\right)$
$\therefore$ Area of triangle $==\frac{2}{\sqrt{3}}$
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