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 $3$ with centre in the first quadrant. If $A$ is one of the points of contact, find the length of $OA$.

Answer 1

$\tan 2\theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{1}{3}$
where $a=2,b=1,h=-3/2$
or $\frac{2t}{1-t^2}=\frac{1}{3}$ where $t=\tan \theta$
or $t^2+6t+1=0$
$\therefore \tan \theta =-3\pm \sqrt{10}$
Since $\theta <{90}^{\circ }\therefore \tan \theta =+ve$ and hence
we choose $\tan \theta =-3+\sqrt{10}(+iv)$
Now from the figure
$OA=3\cos \theta =\frac{3}{\sqrt{10}-3}=\frac{3(\sqrt{10}+3)}{1}$.