You could of course, assuming the actual radii are not important, let the smallest circle vanish to a point, at which point it becomes clear that the radii of the other two circles which must be 65 and 33, become the hypotenuse and one leg of a right triangle, with x as the other leg. x is therefore 56.

If you want to be more rigorous, let the radii be a<b<c.

33^2 = b^2 – a^2

65^2 = c^2 – a^2

x^2 = c^2 – b^2

65^2 – 33^2 = c^2 – a^2 – b^2 +2^2 = c^2 – b^2

x^2 = 65^2 – 33^2

x = 56