Super Smash Ball

Take the two chords as coordinate axes. The centre of the circle lies on the intersection of the perpendicular bisectors of the two chords.

The circle intersects the horizontal axis at $(-a,0)$ and $(b,0)$ . The midpoint of this chord is therefore at $\left(\frac{a+b}{2},0\right)$ , ie $(25,0)$ ; so its perpendicular bisector has equation $x=25$ . Similarly, the perpendicular bisector of the vertical chord is $y=39$ , so the centre of the circle has coordinates $(25,39)$ .

Now, any circle with this centre big enough to intersect the axes will satisfy the conditions $b-a=50$ and $d-c=78$ , so we need to use the fact that $a,b,c,d$ and $r$ (the radius of the circle) are integers.

The equation of the circle is $(x-25)^2+(y-39)^2=r^2$ . Since the point $(0,-c)$ is on the circle, $25^2+(c+39)^2=r^2$

This is a Pythagorean triple! So, for some positive integers $m,n,t$ with $m$ and $n$ coprime, $t\left(m^2-n^2\right)=25;\;\;\;2tmn=c+39;\;\;\;t\left(m^2+n^2\right)=r^2$

We only need the first of these, ie $t\left(m^2-n^2\right)=25$ to see there are just two possible solutions; $(m,n,t)=(3,2,5)$ which gives $r=65$ or $(m,n,t)=(13,12,1)$ , giving $r=313$ .

In the same way, since the point $(-a,0)$ lies on the circle, $(a+25)^2+39^2=r^2$

Now, $313^2-39^2=96448$ , which is not a square number. But $65^2-39^2=2704=52^2$ , so we find $r=\boxed{65}$ .

The rest of the lengths are $(a,b,c,d)=(27,77,21,99)$ .

From the intersecting chords theorem , $ab = cd$ , and from this theorem we know that $a^2 + b^2 + c^2 + d^2 = 4R^2$ .

Now $ab = cd$ rearranges to $b = \frac{cd}{a}$ and $b - a = 50$ rearranges to $b = a + 50$ , so $b = \frac{cd}{a} = a + 50$ , which solves to $a = \sqrt{cd + 625} - 25$ , so that $b = a + 50 = \sqrt{cd + 625} + 25$

Also, $d - c = 78$ rearranges to $d = c + 78$ , so that $a = \sqrt{cd + 625} - 25 = \sqrt{c^2 + 78c + 625} - 25$ and $b = \sqrt{cd + 625} + 25 = \sqrt{c^2 + 78c + 625} + 25$ .

Substituting $a = \sqrt{c^2 + 78c + 625} - 25$ , $b = \sqrt{c^2 + 78c + 625} + 25$ , and $d = c + 78$ into $a^2 + b^2 + c^2 + d^2 = 4R^2$ gives $(\sqrt{c^2 + 78c + 625} - 25)^2 + (\sqrt{c^2 + 78c + 625} + 25)^2 + c^2 + (c + 78)^2 = 4R^2$ , which rearranges to $R^2 - (c + 39)^2 = 625$ .

Letting $k = c + 39$ , we have $R^2 - k^2 = 625$ for integers $R$ and $k$ , or $(R - k)(R + k) = 625$

Out of the factors of $625$ , $R - k$ can be either $1$ or $5$ .

If $R - k = 1$ , then $R + k = 625$ , $R = 313$ , $k = 312$ , $c = 273$ , and $a = 8\sqrt{1507} - 25$ , a non-integer. Therefore, $R - k$ cannot be $1$ and must be $5$ .

If $R - k = 5$ , then $R + k = 125$ , $R = 65$ , $k = 60$ , $a = 27$ , $b = 77$ , $c = 21$ , and $d = 99$ , the only integer solution.

Therefore, $R = \boxed{65}$ .