jmc

The equation of a circle centered at $(h,k)$ with radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$, so the equation of the circle is $$(x-5{)}^2+(y-15{)}^2=130.$$ Since $x=2y$, we substitute to find $$(2y-5{)}^2+(y-15{)}^2=130.$$ Expanding the left-hand side and subtracting 130 from both sides, this equation becomes $$5y^2-50y+120=0.$$ The left-hand side of this equation factors as $5(y-6)(y-4)$, so $y=6$ and $y=4$ are the two possible $y$-coordinates. Therefore, the possible $x$-coordinates are 12 and 8, the largest of which is $\boxed{12}$.