Question

A pair of tangents are drawn from origin to the circle $x^2+y^2-8x+12=0$. Another tangent is drawn to this circle such that this circle will become the incircle of the triangle which is formed by these three tangents. What will be the area of the triangle formed?

• A. $12$
• B. $12\sqrt{3}$
• C. $16\sqrt{3}$
• D. $16$

Answer 1

The correct option is B $12\sqrt{3}$


$\sin \theta =\frac{2}{4}\Rightarrow \theta =\frac{\pi }{6}$
From the figure we can see that
$2\theta ={60}^{\circ }$ and $OA=OB$
So the $△OAB$ will be equilateral,
$\frac{AD}{OD}=\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow AD=\frac{6}{\sqrt{3}}=2\sqrt{3}\Rightarrow AB=2AD=4\sqrt{3}$
Area of $△OAB$
$=\frac{1}{2}\times OD\times AB=12\sqrt{3}$