Question

If the pair of straight lines $\sqrt{3}xy-x^2=0$ is tangent to the circle at $P$ and $Q$ from origin $O$ such that area of the smaller sector formed by $CP$ and $CQ$ is $3\pi$ sq. unit, where $C$ is the centre of circle, then $OP$ equals to

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

Answer 1

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

$\sqrt{3}xy-x^2=0$
$\Rightarrow x(\sqrt{3}y-x)=0$
$\Rightarrow x=0$ or $y=\frac{1}{\sqrt{3}}x$

$\tan {30}^{\circ }=\frac{r}{OP}$
$\Rightarrow OP=r\sqrt{3}$
Area of sector $=\frac{1}{2}{r}^2\theta$
$\Rightarrow \frac{1}{2}\times r^2\times \frac{2\pi }{3}=3\pi$
$\Rightarrow r=3$
$\therefore OP=3\sqrt{3}$ units