Question

Area of the triangle formed by the tangents drawn from $(2,2\sqrt{3})$ to the circle $x^2+y^2=4$ and the chord of contact joining the tangency points is

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $4$
• D. $3\sqrt{3}$

Answer 1

The correct option is D $3\sqrt{3}$



$OP=4\Rightarrow sin\theta =\frac{1}{2}\Rightarrow \theta ={30}^{\circ }$
$\Rightarrow \Delta APB$ is equilateral
Now $AP=\sqrt{16-4}=2\sqrt{3}$
$Area\left(\Delta APB\right)=\frac{\sqrt{3}}{4}\times 12=3\sqrt{3}$