circle

As this question is not solvable by plugging in numbers to a hand-held calculator (or the nasty one on this website), the answer bots unfortunately will not be able to answer your question. However, I will take the liberty to say that the answer is \(5\sqrt2\). Go figure.

We can construct a trapezoid  with bases

CB = sqrt (32) =  4sqrt (2)

And  AD = sqrt (18) = 3 sqrt (2)

Let C = (-4sqrt (2) / 2 , 0) =  (-2sqrt (2),0)  = (-sqrt (8) , 0)

And by symmetry,  B =(sqrt (8) , 0)

Note that DB =   sqrt ( 3^2 + 4^2)  = 5

Construct circles  with radius = 5  at C , B

The equation for the circle at B =   (x -sqrt 8)^2 + y^2   = 25 

D will have the coordinates   ( 3sqrt (2)/2 , y)  =  (3/sqrt (2)  ,y)

And by symmetry A = ( -3/sqrt (2) , y)

We can  find the y coordinate for D   by solving this  for  y

(3/sqrt2 - sqrt 8)^2  + y^2  = 25     ⇒   y =  7/sqrt 2

So D = (3/sqrt (2) , 7/sqrt (2) )

The center of the circle through ABCD    =  E = (0, y)

We can find y by equating  the distance  from E to B  and  from E to D

So

(sqrt (8) - 0)^2 + y^2 =  ( 3/sqrt 2 - 0)^2 + (y - 7/sqrt 2)^2  

Solving this for  y  produces  3/sqrt (2)

So  E = (0, 3/sqrt (2))

The radius of the circle  can  be  found  as EB =   sqrt [  (sqrt (8) - 0)^2  + (3/sqrt (2) - 0)^2 ] =

sqrt [ 8  + 9/2]  = sqrt [ 25 / 2 ]  =   5 / sqrt (2)

So....the diameter  =  2 * 5 / sqrt (2) =  5 (2 / sqrt (2)  = 5sqrt (2)   { as Holtran found !!! }