there is a piont A outside the circle.from A,TWO SECANTS ARE DRAWN,ONE OF WHICH PASSES THROUGH CENTRE OF CIRCLE whose radius is R and other is at a distance of R/2 from centre.Find the area of circular region enclosed by 2 secants.

The circle is divided into 3 regions- a semicircle of area pR²/2, the region whose area is sought, and, a region of a circle of radius R, bounded by ,a chord whose mid point is R/2 from the centre, and the smaller arc of the circle. Consider a chord whose midpoint is between R/2 and R away from the centre. Call this distance ‘r’. It is easy to see that the area of this region is equal to a =  ∫2√(R²-r²) dr from r=R/2 to R . Let r = Rcos(theta). a = ∫2R² sin²(theta)d(theta).from theta = zero to (pi/3).  sin²(theta) = (1-cos2(theta))/2

Thus a =   R²((p/3) – ( √3/4)). So the area which the problem asks is

pR²/2 - R²((p/3) – ( √3/4)) = R²((π/6) +( √3/4))

p=π everywhere