Let m.n be two points on the radius op of the ball 0 and NP = Mn = OM. Make three circles that are perpendicular to the surface of OP through m.n.o. respectively, then the area ratio of these circles is

Let m.n be two points on the radius op of the ball 0 and NP = Mn = OM. Make three circles that are perpendicular to the surface of OP through m.n.o. respectively, then the area ratio of these circles is

Let OP = R, then PN = nm = Mo = R / 3,
Through O, m and N, make the vertical truncated surface of OP respectively, and get the circle O, m and N, whose radii are R, R √ (8 / 9) and R √ (5 / 9), respectively,
Its area ratio is 1: (8 / 9): (5 / 9) = 9:8:5

(Nantong, 2011) as shown in the figure, ⊙ O's chord AB = 8, M is the midpoint of AB, and OM = 3, then the radius of ⊙ o is equal to ()
A. 8B. 4C. 10D. 5

Connecting OA, ∵ m is the midpoint of AB, ∵ om ⊥ AB, and am = 4. In the right angle △ OAM, OA = AM2 + om2 = 5, so D is selected