## HOW TO FIND AREA BOUNDED BY THREE LINES AND CIRCLES , AREA UNDER CURVES BY INTEGRATION METHOD

How to find common area of three lines and one circles which are intersecting at different points with the help of an example.

##
**Given Lines and Curves**

Consider one circle and three lines whose equations are given below( x

^{ }- 1 )

^{2 }+ y

^{2 }= 1

^{2 }

^{ ......................(1)}

^{}y = x .....................(2)

y = -√3 (x-2) .....................(3)

y = 0 .....................(4)

Let us draw these lines and circle in coordinate planes, We can compare the equation of circle with standard form of circle to find the coordinate of centre of the circle is (1,0) and radius of both the circles is 1.

## How To Draw Figure

1st of all check whether these lines intersect with circle or not . And if these lines intersect with each other or with circle then what are their coordinates of points of intersections.

## Solve (1) and (2)

Putting the value of 'y' from (2) in equation (1), we get

( x

^{ }- 1 )^{2 }+ x^{2 }= 1^{2 }^{ }
⇒ x

^{2 }+ 1^{2}-2×1×x +x^{2 }= 1^{2 }^{ }
⇒ -2x +x

⇒ 2x(-1+x) = 0^{2 }= 0
Either 2x= 0 or (-1+x) = 0

⇒ x = 0 or x = 1

Now putting the values of x in (2) we get

x = 0 when x = 0 and y = 1 when x = 1

Therefore points of intersection of (1) and (2) are

**O( 0 , 0 ) and A( 1, 1 )**

## Solve (1) and (3)

Putting the value of 'y' from (3) in equation (1), we get

( x

^{ }- 1 )^{2 }+ [-√3 (x-2)]^{2 }= 1^{2 }
⇒ x

^{ }^{2 }+ 1^{ }^{2 }- 2x + 3 (x-2)^{2 }= 1
⇒ x

^{ }^{2 }+ 1- 2x + 3 [ x^{ }^{2}^{ }+4 - 4x] =1
⇒ x

^{ }^{2 }+ 1- 2x + 3x^{ }^{2}^{ }+ 12 - 12x -1 = 0
⇒ 4x

^{ }^{2}^{ }- 14x +12 = 0
⇒ 2x

^{ }^{2}^{ }- 7x + 6 = 0 , By Factorisation Method
⇒ 2x

^{ }^{2}^{ }- 4x - 3x + 6 = 0
⇒ 2x(x-2) -3( x - 2) = 0

⇒ (x-2)( 2x - 3) = 0

Either (x-2) = 0 or ( 2x - 3) = 0

x = 2 and x = 3/2

To find the values of y , put both the values of 'x' in (3) .i.e. in y = -√3 (x-2)

when x = 2 , then y = 0

and x = 3/2 , then y = √3 /2

Therefore points of intersection of (1) and (3) are

**C( 2 , 0 ) and B(**3/2

**,**√3 /2

**)**

## Solve (2) and (3)

## Same problem with the help of this Video ⇊

##
How to Find Required Area

Required Area = Shaded Area = Area of Î”OAL + Area of Curve ABMLA+ Area of Î” BCM

After simplification , we get

**My Previous Posts**

**How to find area of the circles which is interior to the parabola**

**How to find common area of two parabolas.**

## Final words

^{}

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