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} $.