If you’re familiar with the intersecting chords theorem, we will make use of it here. However the chords we will use actually intersect outside of the circle. Our point is the lower left vertex.

Looking at the diagonal line across the hexagon, its total length is 4x (since a regular hexagon is simply six equilateral triangles), therefore the length to the far side of the circle is 3x, and we are given the distance to the near side of the circle: 1.

The other line going through the lower left vertex is the base of the hexagon, which is tangent to the circle. In this case the distance to the nearside and the distance to the far side are both x.

By the intersecting chords theorem, 1*3x = x*x

3x = x^2

Therefore x=3

To find what the radius of the circle is, draw a line straight up from the tangent point to form another pair of intersecting chords with the main diagonal. The main diagonal within the circle will be split into lengths of 5 and 3, and the vertical line will have lengths of sqrt(27) (half the height of the hexagon) and (2r – sqrt(27)). This eventually works out as the radius r = 7/sqrt(3) = 4.041…