Question 1120521
*[illustration Fig]

since CE=3, EF=7, FD=2
the radius of circle O is (3+7+2)/2=6
that is to say OC=OD=OA=6.
Therefore, EH=3, OF=4.

since O' is the center of the smaller circle(circle O')
O'A=O'B=O'E=O'F.
hence EH=HF=1/2EF=3.5
So OH=5.

Assume AB=x. O'A=x/2.
In right triangle AOO': OO'^2+(x/2)^2=6^2.
In right triangle O'OH: OO'^2=O'H^2+(0.5)^2
In right triangle O'HF: O'H^2+(3.5)^2=(x/2)^2.
Solve the equation set for x(as you can see OO' and O'H can be eliminated through substitution)
And then you get x. square it and you obtain AB^2.

Enjoy!