Using the fact that an angle in a semi-circle is always a right angle, we can see why the given circle passes through the midpoints shown. Similarly we can see why the smaller circle in the second figure also passes through midpoints of the heptagon. Now we only need to find the diameter of that circle. I only give you an overview of the solution for you to fill in the gaps if you desire.

We can find angle ACB easily enough, and since triangle ABC is isosceles, we can find the length of AC.

Triangle ABC is similar to triangle ACD, therefore AB:AC = AC:CD, so we know length CD.

By examining the angles triangle CDE is isosceles, therefore CE = CD.

Since the heptagon is regular, we know the angles within isosceles triangle CEF, so we can know the length of CF.

Moving to the next figure, GH = CE, and IJ = CF.

GHIJ and IJKL are similar trapezoids (sides and diagonals parallel between the two figures) therefore GH:IJ = IJ:KL.

Therefore if we know AB = 6296, we also know that KL = 1247 (to the nearest integer, 1247.000015… to a greater degree of accuracy).

If you’re interested, the exact ratio between AB and KL is:

(1+2sin(270/7))^2