ABC is a right triangle, so the hypotenuse AC is the diameter of its small circle.

It is easy to know OD⊥AC by connecting the O point of the center of the ball with the D point of AC.

D is the center of the small circle, so OD⊥△ABC is at point O,

△ AOC passes through the center of the circle, so the circle where△ AOC is located is the great circle of the ball, and OA is the radius of the ball.

SA⊥△ABC, the small circle of SA⊥, the small circle of SAC⊥,

There is also an OD⊥ small circle, and D belongs to the face capsule. In the same way, there must be an OD belonging to the face capsule.

In the great circle SAC, SA⊥AC, so SC must be the diameter of the great circle, that is, the diameter of the ball, so SC passes through the O point, so =OC.

SC again? =SA？ +AC？ , ∴SC=2, so there is a radius = 1,

So the surface area of the ball O =4π× 1? =4π。