Let the centre of the circle be $(h,k)$. Since it lies on the $x$-axis, $k=0$, so the centre is $(h,0)$. The circle passes through $(3,4)$ and $(1,0)$. The distance from the centre to both points must be equal to the radius, $r$. Thus, $(3-h)^2 + (4-0)^2 = r^2$ and $(1-h)^2 + (0-0)^2 = r^2$. Equating these: $(3-h)^2 + 16 = (1-h)^2$. This expands to $9 - 6h + h^2 + 16 = 1 - 2h + h^2$. Simplifying yields $25 - 6h = 1 - 2h$, which means $24 = 4h \implies h = 6$. The centre is $(6,0)$. Now, calculate the radius from $(1,0)$ to $(6,0)$: $r = \sqrt{(1-6)^2 + (0-0)^2} = \sqrt{(-5)^2} = \sqrt{25} = 5$. The radius of the circle is $5$.