Question

Let $2x^2+y^2-3xy=0$ be the equation of a pair of tangents drawn from the origin $O$ to a circle of radius $3$ units with centre in the first quadrant. If $A$ is one of the points of contact, then the length of $OA$ is

• A. $3+\sqrt{10}$ units
• B. $9(1+\sqrt{10})$ units
• C. $3(3+\sqrt{10})$ units
• D. $9+\sqrt{10}$ units

Answer 1

The correct option is C $3(3+\sqrt{10})$ units

$y^2-3xy+2x^2=0$
Let $\theta$ be the angle between the two lines. Then,
$\tan \theta =\frac{2\sqrt{\left(9/4\right)-2}}{1+2}=\frac{1}{3}$
but from angle between the formulae
$\tan \theta =\frac{2Lr}{L^2-r^2},$
Hence, $\frac{1}{3}=\frac{6L}{L^2-9}$
$\Rightarrow L^2-18L-9=0$
$\Rightarrow L=9\pm \sqrt{90}$
As $L$ can't be negative,
$L=9+\sqrt{90}$
$\Rightarrow L=3(3+\sqrt{10})$
$\therefore OA=L=3(3+\sqrt{10})$ units