CAT 2022 Slot 3 Paper

BC is the diameter of circle C2 so we can say that ∠BAC=90∘ as angle in the semi circle is 90°
Therefore overlapping area =

1

2

(Area‌of‌circle‌C‌2)+Area‌of‌the‌minor‌sector‌made‌be‌BC in C1
AB=AC=8‌cm and as ∠BAC=90°, so we can conclude that BC=8√2‌cm
Radius of C2= Half of length of BC=4√2‌cm
Area of C2=π(4√2)2=32πcm2
A is the Centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8‌cm
Area of the minor sector made be BC in C1=

1

4

(Area‌of‌circle‌C‌1)−Area‌of‌triangle ABC=

1

4

π(8)2−(

1

2

×8×8)=16π−32cm2
Therefore,
Overlapping area between the two circles =

1

2

(Area‌of‌circle‌C‌2)+Area‌of‌the‌minor‌sector‌made‌be‌BC in C1
=

1

2

(32π)+(16π−32)=32(π−1)cm2