Question

Let the tangents at two points $A$ and $B$ on the circle $x^2+y^2-4x+3=0$ meet at origin $O(0,0)$. Then the area of the triangle $OAB$ is

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

Answer 1

The correct option is A $\frac{3\sqrt{3}}{4}$

$x^2+y^2-4x+3=0$
$\Rightarrow (x-2{)}^2+y^2=1$

$AO=\sqrt{(OC{)}^2-(AC{)}^2}$
$=\sqrt{4-1}=\sqrt{3}$
$\sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6}$
Also, $AO=BO$
Area of $\Delta OAB=\frac{1}{2}\cdot OA\cdot OB\sin {60}^{\circ }$
$=\frac{1}{2}\times \sqrt{3}\cdot \sqrt{3}\cdot \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}$