To find area enclosed by
x2 + y2 = 9 (i)
(x – 3)2 + y2 = 9 (ii)
On solving equation (i) and (ii) we get,
x = \(\frac{3}{2}\) and y = \(\pm \frac{3\sqrt{3}}{2}\)
Equation (i) represents a circle with centre (0, 0) and meets axes at (±3, 0),(0, ±3).
Equation (ii) is a circle with centre (3, 0) and meets axe at (0, 0), (6, 0).
They intersect each other at \(\left(\frac{3}{2},\frac{3\sqrt{3}}{2} \right)\) and \(\left(\frac{3}{2},-\frac{3\sqrt{3}}{2} \right)\).
These are shown in the graph below:
Required area = Region OABCO
= 2(Region OBCO)
= 2(Region ODCO + Region DBCD)
The area of the region enclosed between the two curves curves x2 + y2 = 9 and (x – 3)2 + y2 = 9 is \(\left(6\pi - \frac{9\sqrt{3}}{2} \right)\) sq. units.
