Question: Two circles with radius 2R and √2R intersect each other at points A and B. The centers of both the circles are on the same side of AB. O is the center of the bigger circle and ∠AOB is 60°. Find the area of the common region between two circles.
Explanation:
Let O is the center of the bigger circle. So, OA = 2R. Let P be the center of the smaller circle. So, PA = R√2.
In ∆OAM, ∠AOM = 30° and OA = 2R, so AM =OA × Sin 30 = 2R/2 = R. Also, OM = R√3
In ∆PAM, let ∠APM = θ and PA = R√2 and AM = R, so Sin θ = AM/PA = R/ R√2 = 1/√2, so θ = 45°
Therefore ∠APB = 2∠APM = 2 × 45 = 90°. Also, PM = PA × Cos 45 = R√2 × Cos 45 = R
Common area between the two circles is the area of the smaller circle minus the area of the shaded region.
Area of the shaded region = Area of quadrant APB + Area ∆OPA + Area ∆OPB – Area of sector AOB
Area of quadrant APB = π × (R√2)2 /4
Area ∆OPA + Area ∆OPB = Area ∆AOB – Area ∆PAB = [(1/2) × OM × AB) – [(1/2) × PM × AB]
OM = R√3 and AB = AM × 2 = R × 2 = 2R
So, Area ∆OPA + Area ∆OPB = [(1/2) × R√3 × 2R) – [(1/2) × R × 2R] = (√3 – 1)R2
Area of sector AOB = (π/6)(2R)2
Hence, Area of the shaded region = [π × (R√2)2 /4] + [(√3 – 1)R2 ] +[(π/6)(2R)2 ]
= [(√3 – 1)R2 ] – [πR2 /6]
Common area between the two circles = Area of the smaller circle – Area of the shaded region
= π(R√2)2 – {[(√3 – 1)R2 ] – [πR2 /6] = (13π/6) + 1 – √3) R2
Hence, option (c).