Let ON⊥ and OA 45? A straight line intersects the circle O in B. Because M is the midpoint of OA, OM = 1/2oa, so ON = √ 2/4oa = √ 2/4r.

This figure is a cross-sectional view of the radius OA in the circle O. Obviously, the line segment cut by MN and the circle is the diameter of the new circle cut by plane. Then the distance from the new circle to O is √2/4R, and the radius of the new circle is r = √( r the square of r-the square of √ 2/4r) = √ 14/4r, and r = 2 √ 2/3 is obtained from π r 2 = π 7/9. The surface area of ball O is 32π/9.