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 radius is equal to the distance from $(1,2)$ to the appropriate angle bisector of the two tangent lines. These bisectors are $${2x-y\over\sqrt5}=\pm{3x-y\over\sqrt{10}}.$$ Choose the one for which the circle will be in the first quadrant.
