I'm trying to find a point on which to place a fourth circle which intersects all three existing circles- of unbounded integer radius
I'm assuming you mean that the fourth circle intersects the centres of the 3 other circles (otherwise you could pretty much draw ANY circle).
This will in many cases be impossible EVEN IF you manage to get an intersection of three circles.
Imagine C1 at (0,0) with r=3, C2 at (0,1) with r=2, C3 at (0,2) with r=1.
They intersect (well not truly intersect but that's not the problem here) at (0,3), but if you then draw a circle at (0,3) with some radius, the centres of C1,2,3 will never all be on the arc of C4.
So what you are stating is no requirement
for the solution of your problem, though it is a nice bonus. This bonus can only be achieved if C1,2,3 all have the same radius.
However, it is possible to have a C4 intersecting the centres of C1,2,3 at the same time. I just can't get all of the equations on here out of my head. (That's a shame for a mathematician).
Now back to the matter at hand.
You not only want them to intersect (easily done), you also want them all to have r∈ℤ. There is only one solution, due to the location of the centre of C3 (0.5;sqrt(3/4)).
I found this solution while searching for formulas in the situation of r1=r2. Then the arcs of C1 intersect at (0.5;sqrt(n^2-0.5^2)). If you discard the requirement that n should be an integer and take n=0,5 then you'll find them intersecting at the x-axis (which is just right because since the distance between C1 and C2 equals 1 and take the radii of 0.5, they'll meet in the middle at the x axis because C1 and C2 lie on the x-axis).
Now, n must be an integer, so it's 0,1,2,... (integers are a selection of Z
so 0 and negative numbers can join in here, but negative numbers aren't an option because a radius is nonnegative).
If we take n=1, then we get the intersection at (0.5,sqrt(3/4)), which is the centre of C3.
So take r3=0 and you have solved the problem. Additionally, Dots can be named circles with r=0 so C3 is still a perfectly fine circle. The other intersection is at (0.5;-sqrt(3/4)) so C3 should have r3=2sqrt(3/4) to join in that intersection but then r3 is no integer)
Taking n>1, we'll get the y-coordinate of the intersection of C1 and C2 to be at sqrt(n^2-1/4). This will be 15/4, 35/4, 63/4, 99/4, 143/4,... for n=2,3,4,5,6...
These will never give an integer as coordinate.
That means that the radius of C3 will be (y-coordinate minus sqrt(3/4)) or (minus y-coordinate + 2sqrt(3/4)) The latter for the intersection in the negative area. Also these will never give an integer.
That means that the only solution for r1=r2 will be (r1,r2,r3)=(1,1,0)
It gets a lot more complicated for having r1 != r2, but here you can still work with symmetry because your figure is symmetrical.
If r1 > r2, the intersections will move to the right, but also the y-coordinate will change (it will increase first - positive region- , as r1 increases, but then it'll decrease - the y-coordinate will be zero if r1=r2+1 leaving only 1 intersection).
If r1 < r2 then you'll have exactly the same, but with the intersections moving to the left.
TL:DR - there is only one solution, that is (r1,r2,r3)=(1,1,0)