Question

A pair of tangents $OA,OB$ $(O$ is origin$)$ is drawn to a circle whose centre is $C$ and radius is $3$ units. If the combined equation of $OA$ and $OB$ is $2x^2-3xy+y^2=0,$ then the area (in sq. units) of the quadrilateral $OACB$ is equal to

• A. $3(3+\sqrt{10})$
• B. $9(3+\sqrt{10})$
• C. $6(3+\sqrt{10})$
• D. $12(3+\sqrt{10})$

Answer 1

The correct option is B $9(3+\sqrt{10})$


Combined equation of $OA$ and $OB$ is $2x^2-3xy+y^2=0$

Let $\theta$ be the angle between $OA$ and $OB.$
Acute angle between the pair of straight lines represented by $a{x}^2+2hxy+b^2=0$ is
$\tan \theta =\frac{2\sqrt{h^2-ab}}{|a+b|}\tan \theta =\frac{2\sqrt{\frac{9}{4}-2}}{3}=\frac{1}{3}\Rightarrow \frac{2\tan \theta /2}{1-{\tan }^2\theta /2}=\frac{1}{3}\Rightarrow \frac{2\left(3/x\right)}{1-\left(9/x^2\right)}=\frac{1}{3}\Rightarrow x^2-18x-9=0\Rightarrow x=9+3\sqrt{10}$

Area of $△OAC$
$=\frac{1}{2}\cdot x\cdot 3=\frac{3}{2}(9+3\sqrt{10})$
$\therefore$ Area of quadrilateral $OACB$
$=3(9+3\sqrt{10})=9(3+\sqrt{10})$ sq. units