We can start off by saying that c>=b>=a, and worry about reflections and rotations later.

First, we’ll deal with situations where some of the lengths are equal. Clearly they can’t all be equal, as the point would be the centre of the circle, not on it. If a=b, then c would be a diameter of the circle, which is 49*(2/rt(3)) ~ 56.5, not an integer. If b=c, then a=0 and we have a vertex of the original triangle. There are three vertices, so already three points which satisfy the question.

We can now make use of a couple of facts about cyclic quadrilaterals: by Ptolemy’s theorem 49c = 49a+49b, so therefore a+b is always equal to c, which is convenient since if a and b are both integers, c will be too. Because opposite angles of a cyclic quadrilateral sum to 180, the angle between a and b is 120 degrees.

Now using the fact that cos(120) = -0.5, we can use the cosine rule to find the following equation:

49^2 = a^2 + b^2 + ab

If we add ab to both sides, then factorise the right-hand side we get:

49^2 + ab = (a+b)^2

a+b=c must be between 49 and the diameter ~56.5, so there are only 7 possible integer values to try, and each will give a different value for ab

(a+b,ab) = (50,99),(51,200),(52,303),(53,408),(54,515),(55,624),(56,735)

Only the last two of these have solutions in the integers:

(a,b,c) = (16,39,55) and (21,35,56).

In our diagram a is the distance from the top vertex, b from the left vertex and c from the right vertex, but these vertices can be permuted in 3! = 6 ways, so these two numerical solutions represent 12 points on the circle. Along with the three vertex points themselves there are therefore 15 points on the circle that are an integer distance from all three vertices of the triangle.