Question

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

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

Answer 1

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

Pair of straight lines:

$x(y\sqrt{3}-x)=0$

$\Rightarrow x=0$ and $y=\frac{x}{\sqrt{3}}$

$\Rightarrow y=x\tan {30}^0$

The straight line makes an angle of ${30}^0$ with the $x-$axis.

Hence, angle $QOP={90}^0-{30}^0={60}^0$

Now $QCOP$ is a cyclic quadrilateral.

Hence, angle $QCP={120}^0$

Therefore, area of smaller sector $QCP$

$\frac{{120}^0}{{360}^0}\pi r^2=3\pi$

$\Rightarrow r^2=9$

$\Rightarrow r=3$

Hence, radius of the circle is $3$.

$CO$ acts as the angle bisector,

$\Rightarrow \angle COP={30}^0$

$OP=r\cot {30}^0$

$\Rightarrow OP=3\sqrt{3}$ units.

Hence, option B.