Question

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

• A. $\frac{\left(3\sqrt{3}\right)}{2}$
• B. $3\sqrt{3}$
• C. $3$
• D. $\sqrt{3}$

Answer 1

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

Given, pair of tangent $xy\sqrt{3}-x^2=0$ from origin

Area of sector $=3\pi$

Let the equation of the circle

$(x-h{)}^2+(y-k{)}^2=r^2$

Two tangents are $x=0,y=\frac{1}{\sqrt{3}}x$

Angle between two tangent $=\frac{\pi }{3}$

Angle $PCQ={120}^o=\frac{2\pi }{3}$

Area of sector $=3\pi$

$\frac{2\pi }{3\times 2\pi }\left(\pi r^2\right)=3\pi r=3units$

In $△COP$

$\tan {30}^o=\frac{CP}{PO}\Rightarrow PO=\frac{CP}{\tan 30}=3\sqrt{3}units$