Capital city O is located mid-way between airforce base A and naval base B?

Question

Enemy air force base C is located at equal distance from point O: AO = BO = CO, and has two types of long range bombers deployed:
Type-Ca(aimed to attack base A) has maximum range CA.
Type-Cb(aimed to attack base B) has maximum range CB.
http://i2.tinypic.com/34odp3a.jpg

Areas of surface of the planet covered by each type of bombers are Sa and Sb respectively.

Long range fighters deployed at AFB A have maximum range AB (just sufficient to defend the naval base).

What is area of planet surface S is covered by long range fighters deployed at A?

Answer 1

*EDIT* I'm correcting an earlier error

Let S be the spherical circle (on the planet's surface), and S' be the corresponding planar circle which is the image of S on the plane passing through the endpoints of S.

Let R = radius of planet
r = radius of S (curved distance)
r' = radius of S' (planar distance)
A = area of S (curved surface)
A' = area of S' (planar surface)
Let 2θ = angle formed by S at the center of the planet

Doing some spherical geometry, we can show that
r = Rθ
r' = Rsinθ
A = 2πR^2 (1 - cosθ)
A' = πR^2 sin^2 θ

Consider now the planar circle passing through A,B,C centered at O'. A,B,C lies on the planar circle with AB as the diameter. So
(AB)^2 = (AC)^2 + (BC)^2 ... (1)
where AB etc are the straight line distances between the points.

AC' would be the r' for Sa, but AC is neither r nor r'. So we need an expression for AC in terms of AC' or Sa. Again from the geometry, we can show that
AC^2 = 2R^2 * (1 - cosθ)
but 1 - cosθ = A / (2πR^2)
So AC^2 = A/π ...(2)
where A = curved surface area

AC corresponds to Sa, BC corresponds to Sb and AB corresponds to S. So substituting eqn (2) into eqn (1),
AC^2 = Sa/π, BC^2 = Sb/π, AB^2 = S/π
S = Sa + Sb

Amazingly the curved surface areas follow the same relation to each other as the plane areas worked out by the first answerer.

Answer 2

as AB is a diameter in the red circle that its center is O
so the angle ACB is right angle
so AB^2 =AC^2 + BC^2
Sa= π AC^2 ---> AC^2 =Sa /π
Sb =π BC^2 ----> BC^2 = Sb /π
so AB^2 =(Sa+Sb) /π
so the area = π AB^2 = π(Sa+Sb)/π =Sa +Sb