work the radius of circle (C1) must be neither too small nor too large. More precisely, this radius must be between half and double of the radius of (C): if the radius is greater than the diameter of (C), (C1) will not intersect (C); if the radius is shorter than half the radius of (C), point C will be between A and O and (C3) will not intersect (C1). Proof The idea behind the proof is to construct, with compass alone, the length b²/a when lengths a and b are known, and a/2 ≤ b ≤ 2a. In the figure on the right, a circle of radius a is drawn, centred at O; on it a point A is chosen, from which points B and B' can be determined such that AB and AB' have a length of b. Point A' lies opposite A, but does not need to be constructed (it would require a straightedge); similarly point H is the (virtual) intersection of AA' and BB'. Point C can be determined from B and B', using circles of radius b. Triangle ABA' has a right angle at B and BH is perpendicular to AA', so : Therefore, and AC = b²/a. In the above construction of the center, such a configuration appears twice : points A, B and B' are on the circle (C), radius a = r ; AB, AB', BC, and B'C are equal to b = R, so ; points A, D and D' are on the circle of centre C, radius ; DA, D'A, DO, and D'O are equal to b = R, so . Therefore, O is the centre of circle (C). Finding the middle of a given distance or of a line segment Let |AD| be the distance, whose centre is to be found. Two circles (C1) centered at A and (C2) centered at D with radius |AD| meet at B and B'. A circle (C3) centered at B' with radius |B'B| meets the circle (C2) at A'. A circle (C4) centered at A' with radius |A'A| meets the circle (C1) at E and E'. Two circles (C5) centered at E and (C6) centered at E' with radius |EA| meet at A and O. O is the sought center of |AD|. The design principle can also be applied to a line segment . The proof described above is also applicable for this design.