If we draw a line from the origin, directly through the centre of the circle and to the point where the arc and the circle meet, that line is clearly equal to R, but it is also equal to (sqrt(2)+1)*r, where r is the radius of the circle. So we have R = (sqrt(2)+1)*r, and if we square both sides (you’ll see why in just a moment) we get:

R^2 = (2*sqrt(2)+3)*r^2

We can also form a right-angle triangle using the origin and the line of length 1, to get:

R^2 = (2r)^2 + 1 = 4r^2 + 1

So we have two alternative equations for R^2, which we can equate:

(2*sqrt(2)+3)*r^2 = 4r^2 + 1

Making r^2 the subject gives:

r^2 = 1/(2*sqrt(2)–1)

Which we can make the denominator rational by multiplying top and bottom by 2*sqrt(2)+1:

r^2 = (2*sqrt(2)+1)/7

We can find R^2 by using either of the relations we worked out previously:

R^2 = 4(2*sqrt(2)+1)/7 + 1

R^2 = (11+8*sqrt(2))/7

Finally R= sqrt((11+8*sqrt(2))/7) ~ 1.7854

Now for the 3-point solution, let R^2 = x

x = (11+8*sqrt(2))/7

but we would like it to look like the quadratic formula (-b +/- sqrt(b^2-4ac))/2a , so the first thing we need is an even number in the denominator, and we also need to bring the 8 under the square root:

x = (22+sqrt(512))/14

So, a = 7, b = -22, and b^2-4ac = 512, 484 - 28c = 512, therefore c = -1

So the polynomial in x is:

7x^2 – 22x - 1

And back in terms of R is:

7R^4 - 22R^2 - 1

This polynomial has four roots, however two are complex, and one is negative, and so the only positive real root is the value for R that we previously found.