Look before you leap

We first look for all integral points on the circle. Let such a point be $h,k$

$h^2,k^2=(25,25),(1,49),(49,1)$

Thus for each of the 3 pairs, we will get 4 permutations of + and - .[e.g.(25,25) will give $(h,k)$ as $(5,5),(-5,5),(5,-5),(-5,-5)$ ].

That gives us a total of 12 points.

Now the given line can either have one solution or 2 solutions with the circle, i.e. it can either be a chord or a tangent.

For tangents, we have 12 ways we could pick a point for it to be the contact point of the tangent to the circle. ( Each of these points will surely give distinct and rational values of $a,b$ .)

For chords, we have $¹²C_{2} =66$ ways to pick 2 points for the chord to pass through them.

This gives the total pairs = 12 + 66 = 78. But notice that some of the chords above will pass through the origin (0,0), which would give the equation of such a chord in the form $y=mx$ . There is no way to represent $ax+by=1$ in the form of $y=mx$ , thus these lines have to be avoided.

There are 6 such lines, thus the final answer becomes $78-6=72$ .