UPSC NDA Math Model Paper 1

y2=x........(1)
and Circle x2+y2=2x......(2)
⇒x2−2x+1−1+y2=0 ⇒(x−1)2+(y−0)2=1
So, centre of circle (1,0) and radius =1
Put (1) in (2)
x2+x=2x⇒x2−x=0 ⇒x(x−1)=0x=1,0
So, point of intersections are (0,0) and (1,1) Required area = =

1

∫

0

(√1−(x−1)2−√x)dx =[

x−1

2

√1−(x−1)2+

1

2

sin−1(x−1)]01 −[

x 3 2

3 2

]01 =[0+

π

4

]−

2

3

=

π

4

−

2

3

Hence, option (4) is correct.