If pair of tangents to a circle in the first quadrant is $6x^2-5xy+y^2=0$ and if one point of contact is $(1,2)$, find the radius. The tangents are $2x-y=0$ and $3x-y=0$. Let the radius be $r$ and centre be $(h,k)$
$$r=\frac{|3h-k|}{\sqrt {10}}$$
$$r=\frac{|2h-k|}{\sqrt 5}$$
$$(h-1)^2+(k-2)^2=r^2$$
I invested a considerable amount of effort in solving this equations, but to no result. My method was to square all terms to avoid the modulus, but that complicated things adding the $hk$ term. This also led me to believe that there must be a better way to solve this. Can I know how this problem must be approached?

The perpendicular to $y=2x$ throught $(1,2)$ is:
$$y=-\frac{1}{2}x+\frac{5}{2}$$
The line bisector of the two lines $y=2x$ and $y=3x$ is:
$$y=(\sqrt2+1)x$$
The other bisector line is:
$$y=(\sqrt2-1)x$$
but in this case the circunference wouldn't be tangent to either $y=2x$ and $y=3x$ lines.
Now, we have to inresect these two lines, or:
$$(\sqrt2+1)x=-\frac{1}{2}x+\frac{3}{2} \leftrightarrow x=5(3-2\sqrt2) \land y=-5+5\sqrt2$$
From this we can compute:
$$r=\sqrt{(14-10\sqrt2)^2+(-7+5\sqrt2)^2}=\sqrt{(5\sqrt{10}-7\sqrt5)^2}=5\sqrt{10}-7\sqrt5$$
Note that we pass from $(14-10\sqrt2)^2+(-7+5\sqrt2)^2$ to $(5\sqrt{10}-7\sqrt5)^2$, simply calculating the first sum and then solve for $a,b$ this system:
$$\left\{\begin{matrix}
a^2+b^2=495
\\ 2ab=-350\sqrt2
\end{matrix}\right.$$
And the solutions are:
$$a=\pm5\sqrt{10} \land y=\mp7\sqrt{5}$$
