BC is the diameter of circle C2 so we can say thatΒ $$\angle BAC=90^{\circ\ }$$ as angle in the semi circle isΒ $$90^{\circ\ }$$
Therefore overlapping area =Β $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1
AB= AC = 8 cm and asΒ $$\angle BAC=90^{\circ\ }$$, so we can conclude that BC=Β $$8\sqrt{2}$$ cm
Radius of C2 = Half of length ofΒ BC =Β $$4\sqrt{2}$$ cm
Area of C2 =Β $$\pi\left(4\sqrt{2}\right)^2=32\pi$$Β $$cm^2$$
A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm
Area of the minor sector made be BC in C1 =Β $$\frac{1}{4}$$(Area of circle C1) - Area of triangle ABC =Β $$\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$$Β $$cm^2$$
Therefore,
Overlapping area between the two circles= $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1
=Β $$\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$$Β $$cm^2$$