linguistic:voice_change, structural:expansion, stylistic:tone, stylistic:style_transfer
Sagitta

Another proof of this result, which relies only on two chord properties given above, is as follows. Given a chord of length y and with sagitta of length x, since the sagitta intersects the midpoint of the chord, we know that it is a part of a diameter of the circle. Since the diameter is twice the radius, the "missing" part of the diameter is (2 r − x) in length. Using the fact that one part of one chord times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2 r − x) x = (y / 2) 2. Solving for r, we find the required result.