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 center in the first quadrant. if A is one of the points of contact find length of OA?

the given pair of tangents are $2x^2+y^2-3xy=0........\left(1\right)$
$$\begin{gathered} 2x^2-2xy-xy+y^2=0 \\ 2x(x-y)-y(x-y)=0 \\ (x-y)(2x-y)=0 \\ x-y=0 or 2x-y=0 \end{gathered}$$

thus y = x and y =2x are the two tangents of the circle at the points B and A respectively.
let $m_1=1 and m_2=2$ 
let $2\alpha$ be the angle between these two tangents.
let $\angle AOB =2\alpha \Rightarrow \angle AOC=\alpha$
$$\begin{gathered} \tan 2\alpha =\frac{m_2-m_1}{1+m_1{m}_2} \\ \tan 2\alpha =\frac{2-1}{1+2}=\frac{1}{3} \\ \frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{1}{3} \\ 1-{\tan }^2\alpha =6\tan \alpha \\ {\tan }^2\alpha +6\tan \alpha -1=0 \\ \tan \alpha =\frac{-6\pm \sqrt{36+4}}{2}=\frac{-6\pm 2\sqrt{10}}{2} \\ \tan \alpha =-3\pm \sqrt{10} \end{gathered}$$
as $\alpha$ lies between 0 to $\frac{\pi }{4}$, $\tan \alpha =\sqrt{10}-3$
the radius of the circle = 3 cm, AC = 3
$$\begin{gathered} \tan \alpha =\frac{AC}{OA} \\ OA=\frac{AC}{\tan \alpha }=\frac{3}{\sqrt{10}-3} \\ OA=\frac{3}{\sqrt{10}-3}*\frac{\sqrt{10}+3}{\sqrt{10}+3} \\ OA=\frac{3(\sqrt{10}+3)}{10-9}=3\sqrt{10}+9 \end{gathered}$$

hope this helps you.