length of a tangent

Question

The two tangents to a circle are represented by $2x^2-3xy+y^2=0$ . A circle of radius=3 is in first quadrant . "A" is a point of tangency where one of these lines meet.What is length OA where $O$ is origin.$$\text{MY ATTEMPT}$$ So I fixed centre as $(h,k)$ now we know length of tangents from commom point to the circle is same but for that we need the points . So i found out equations of two tangents which are $x=y,x=2y$ Now by distance formula we get $$(0-x)^2+(0-x)^2=(0-x)^2+(0-2x)^2$$ as distance is same . but this gives $x=0$ and thus $y=0$ thus distance is $0$ but answer in my book is $18.5$ how to proceed any hints.

Answer 1

The direction vectors of the tangent lines are $(-1,1)$ and $(1,2)$ resp. Add their unit vectors to achieve a direction vector of their bisector, namely $(5\sqrt2+2\sqrt5,5\sqrt2+4\sqrt5)$. Now plug the bisectors equation in the Hesse form of $y=x$, this term has to equal $3$. Thus you'll find the center of the circle $(3\sqrt2+3\sqrt5,6\sqrt2+3\sqrt5)$. Convince yourself the the searched distance is the length of $(3\sqrt2+3\sqrt5-\sqrt3,6\sqrt2+3\sqrt5-\sqrt3)$.

Alternatively, compute the squared length of $(3\sqrt2+3\sqrt5,6\sqrt2+3\sqrt5)$ to be $180+54\sqrt{10}$, hence $x^2+9=180+54\sqrt{10}$.