Cyclics

Let $|\overline {AB}|=25,|\overline {BC}|=39,|\overline {CD}|=52,|\overline {DA}|=60$ . We see that, since $39^2+52^2=65^2,\triangle {BCD}$ is a right angled triangle, right angled at $C$ . Also, since $25^2+60^2=65^2, \triangle {ABD}$ is a right-angled triangle with right angle at $A$ . So the common hypotenuse of the two triangles is the side $\overline {BD}$ of length $65$ . So $\overline {BD}$ is the diameter of the circle with length $\boxed {65}$ .

The radius $R$ of the circumcircle of the cyclic quadrilateral is given by the Parameshvara's circumradius formula :

$R = \frac 14 \sqrt{\frac {(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$

where, $a$ , $b$ , $c$ , and $d$ are the side lengths of the cyclic quadrilateral, and $s = \dfrac {a+b+c+d}2$ , the semiperimeter. Putting in the values of $25$ , $39$ , $52$ , and $60$ for $a$ , $b$ , $c$ , and $d$ , we get $R=32.5$ , hence the diameter of the circumcircle is $\boxed {65}$ .