We draw EF perpendicular to both bases of the trapezoid, bisecting the
trapezoid, which makes EA=1 and FD=4 



Next we draw AG parallel to EF, which divides FD into FG=1 and GD=3,
and by the Pythagorean theorem, AG=4 since triangle AGD is a 3-4-5
right triangle: 



Next, let's sketch in the circle the trapezoid is inscribed in, and
let its center be O, which is a point on EF.



Next we draw x and y axes, so that the CD is along the x-axis, and
EF is along the y-axis.  That makes the point A be (1,4), and D be (4,0),
Then the equation of the circle is 
,

Since O is on the y-axis, O's x-coordinate in h=0. Its y-coordinate is k.

So the equation of the circle is





The points A(1,4) and D(4,0) lie on the circle, so we substitute each
in the equation of the circle:





You finish.  Solve that system by substitution and get 



and



Edwin