Question 1150431
<pre>

{{{drawing(400,200,-6,6,-1,5,
locate(1,4.5,A), locate(-1.1,4.5,B),locate(4,0,D),locate(-4,0,C),
locate(2.5,2.5,5),locate(0,4.5,2), locate(0,0,8),

line(1,4,-1,4), line(-4,0,4,0), line(-1,4,-4,0),line(4,0,1,4) )}}}

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

{{{drawing(400,200,-6,6,-1,5,
locate(1,4.5,A), locate(-1.1,4.5,B),locate(4,0,D),locate(-4,0,C),
locate(2.5,2.5,5),locate(-.1,4.5,E), locate(-.1,0,F),
green(line(0,0,0,4)),locate(.45,4.5,1),locate(1.95,0,4),
line(1,4,-1,4), line(-4,0,4,0), line(-1,4,-4,0),line(4,0,1,4) )}}}

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: 

{{{drawing(400,200,-6,6,-1,5,
locate(1,4.5,A), locate(-1.1,4.5,B),locate(4,0,D),locate(-4,0,C),
locate(2.5,2.5,5),locate(-.1,4.5,E), locate(-.1,0,F),
green(line(0,0,0,4),line(1,0,1,4)),locate(.45,4.5,1),locate(2.1,0,3),
locate(.45,0,1),locate(1,0,G), locate(1,2,4),
line(1,4,-1,4), line(-4,0,4,0), line(-1,4,-4,0),line(4,0,1,4) )}}}

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

{{{drawing(400,200,-6,6,-1,5,
circle(0,.3,.13),

locate(1,4.5,A), locate(-1.1,4.5,B),locate(4,0,D),locate(-4,0,C),
locate(2.5,2.5,5),locate(-.1,4.5,E), locate(-.1,0,F), locate(1,2,4),
green(line(0,0,0,4),line(1,0,1,4)),locate(.45,4.5,1),locate(2.1,0,3),
locate(.45,0,1),locate(1,0,G), circle(0,1/8,5sqrt(41)/8+.06),
locate(-.3,.6,O),
line(1,4,-1,4), line(-4,0,4,0), line(-1,4,-4,0),line(4,0,1,4) )}}}

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 
{{{(x-h)^2+(y-k)^2=r^2}}},

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

{{{x^2+(y-k)^2=r^2}}}

{{{drawing(400,200,-6,6,-1,5,
circle(0,.3,.1),
graph(400,200,-6,6,-1,5), locate(1,2,4),
locate(1,4.5,"A(1,4)"), locate(-1.1,4.5,B),locate(4.2,.7,"D(4,0)"),locate(-4,0,C),
locate(2.5,2.5,5),locate(-.1,4.5,E), locate(-.1,0,F),
green(line(0,0,0,4),line(1,0,1,4)),locate(.45,4.5,1),locate(2.1,0,3),
locate(.45,0,1),locate(1,0,G), circle(0,1/8,5sqrt(41)/8+.06),
line(1,4,-1,4), line(-4,0,4,0), line(-1,4,-4,0),line(4,0,1,4) )}}}

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

{{{system(1^2+(4-k)^2=r^2,4^2+(0-k)^2=r^2)}}}

{{{system(1+(4-k)^2=r^2,16+k^2=r^2)}}}

You finish.  Solve that system by substitution and get 

{{{k = 1/8}}}

and

{{{r = (5sqrt(41))/8}}}

Edwin</pre>