EAMCET Engineering 2010 Solved Paper

Intersection point of two circlesx2+y2=1 . . . (i)
(x−1)2+y2=1 . . . (ii)
is given by
‌(x−1)2+(1−x2)=1
⇒x2+1−2x−x2=0
⇒x=‌

1

2

From Eq. (i), ‌

1

4

+y2=1
y2=1−‌

1

4

⇒y=±‌

√3

2

Point A(‌

1

2

,‌

√3

2

) and C(‌

1

2

,‌

−√3

2

)
So, Area of region OABCO
=2× Area of region OABDO . . . (i0
Area of OABDO
= Area of OADO+ Area of ABDA
‌=

1∕2

∫

0

√1−(x−1)2‌dx+

1

∫

1∕2

√1−x2‌dx
‌=[‌

1

2

⋅(x−1)√1−(x−1)2+‌

1

2

sin‌−1(‌

x−1

1

)]01∕2
‌+[‌

1

2

x√1−x2+‌

1

2

sin‌−1(‌

x

1

)]1∕21
‌=[−‌

1

2

⋅‌

1

2

⋅‌

√3

2

+‌

1

2

sin‌−1(‌

−1

2

)−‌

1

2

⋅0−‌

1

2

sin‌−1(−1)]
‌+[‌

1

2

⋅0+‌

1

2

sin‌−1(1)−‌

1

4

⋅‌

√3

2

−‌

1

2

sin‌−1(‌

1

2

)]
‌=(−‌

√3

8

)−‌

1

2

sin‌−1(‌

1

2

)+‌

1

2

sin‌−1(1)
‌+‌

1

2

sin‌−1(1)−‌

√3

8

−‌

1

2

sin‌−1(‌

1

2

)
‌=sin‌−1(1)−sin‌−1(‌

1

2

)−‌

√3

4

‌=‌

π

2

−‌

π

6

−‌

√3

4

‌=(‌

π

3

−‌

√3

4

)
from Eq. (i),
‌ Area of region ‌OABCO‌=2×(‌

π

3

−‌

√3

4

)
‌=(‌

2π

3

−‌

√3

2

)