(a) The squared radius is \((7-5)^2+(5-2)^2=13\). Therefore \[\boxed{(x-5)^2+(y-2)^2=13}.\]

(b) Substitute the line equation into the circle: \[(x-5)^2+(5x-12)^2=13.\] Expanding and simplifying gives \[26x^2-130x+156=26(x-2)(x-3)=0.\] The intersection points are \((2,0)\) and \((3,5)\). Their distance is \[\boxed{AB=\sqrt{(3-2)^2+(5-0)^2}=\sqrt{26}}.\]