## Wednesday 16 January 2019

### 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 y2 12            ......................(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 x2 12

x2  + 12 -2×1×x +x2 12

-2x +x2 = 0
⇒   2x(-1+x) = 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 12
⇒  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)

is the coordinate of Point of intersection of Line (2) and (3)

## 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

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