# Two intersecting circlesWatch

This discussion is closed.
#1
This is a wierd one I set my self to keep my self awake over the summer but it is too hard for me.

Two identical circles overlap, so that each passes through the other's centre. What is the area in terms of r of the region which both circles ocupy.

a bit like this venn diagram but the circles are closer so that one goes through the other's centre and I want the area of region c on that diagram.

It looks like but I cant be sure that one of the points where the circles meet, and the two centres make an equalateral triangle. That would help.

thanks
0
14 years ago
#2
It's not hard. All you have to know is the formula for the area of a triangle, which is 1/2*a*b*sin c, as well as the area for sectors of a circle.
0
14 years ago
#3
(Original post by sephonline)
It's not hard. All you have to know is the formula for the area of a triangle, which is 1/2*a*b*sin c, as well as the area for sectors of a circle.
yeah, it's not hard.
Area = 4*integral ((-r) --> (-r/2)) of sqrt(r^2 - x^2).
0
14 years ago
#4
Using calculus:

Circle 1: x^2 + y^2 = r^2

Circle 2: x^2 + (y-r)^2 = r^2

Substitute into each other and solve for x and y:

r^2 - y^2 = r^2 - (y-r)^2
y^2 = y^2 -2ry +r^2
y=r/2

therefore x = (sqrt(3)/2)*r and x = -(sqrt(3)/2)*r

Now, we need to only consider the upper half of circle 1 and the lower half of circle 2 so we get the formula's

y1=sqrt(r^2-x^2)
y2=r-sqrt(r^2-x^2)

Then using calculus to find the area under the graphs we get

A = Int{[From -(sqrt(3)/2)*r to (sqrt(3)/2)*r] of (y2 - y1)}

= Int{[From -(sqrt(3)/2)*r to (sqrt(3)/2)*r] of (r - 2*sqrt[r^2-x^2])}

= [rx -x*sqrt[r^2-x^2] +r^2*arcsin(x/r) from (-(sqrt(3)/2)*r to (sqrt(3)/2)*r)]

= [(sqrt(3)/2) - (2/3)*PI]*r^2

I'll try and tidy that up with some nice TeX later.
0
14 years ago
#5
Oh, it's quite long way . Just draw a diagram and see. Area C will be divided by 4 equal parts.
And consider a circle: x^2 + y^2 = r^2, then calculate one part by integral.. then time 4
0
14 years ago
#6
Thats a good point.
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14 years ago
#7
this kind of question is best done using polar coordinates and integration
0
14 years ago
#8
As sephonline said, it's not that hard and surely there's no need to resort to calculus. Not sure the person posing the question will have done integration.

The region comprises two segments, and the area of a segment is just the area of a sector minus a triangle.

The area of a sector subtending an angle theta (radians) is

1/2 r^2 theta

as it's theta/(2pi) of the whole circle's area. [If you've not met radians yet the formula is (angle in degrees/360) pi r^2]

And the angle at the centre is 2pi/3 as half the angle is subtended by a right-angled triangle with hypoteneuse r and adjacent r/2.
0
14 years ago
#9
Area of C = (rÂ²)(2π/3 - √3/4)

it makes 4 segments and 2 equilateral triangles...

area of traingles =

(1/2)(rÂ²)(sin 60)
(1/2)(rÂ²)(√3/2)

area of both triangles is twice the above results :
(rÂ²)(√3/2)

area of sectors is given by (1/2)rÂ²(theta) where theta is in radians so area of each segment is area of sector minus area of triangle so total area of segments is:

2(rÂ²π/3 - rÂ²√3/2)

Adding the areas of two triangles and this results gives the required result
0
14 years ago
#10
Area = [(2/3)π - √(3/4)]rÂ²

Yep.
0
14 years ago
#11
Nope, root(3)/2 is correct.
0
14 years ago
#12
(Original post by AntiMagicMan)
Nope, root(3)/2 is correct.
bah..yes it is
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14 years ago
#13
or root(3/4)
0
14 years ago
#14
Of course, because they are the same thing.
0
#15
yes I know what integration is. I have the alevel.

thanks for that. I skim read your solutions so that I only got an idea of what you did, so I can do it myself and check it with yours later.

thanks
0
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