2 circles A and B with radius a and b units respectively ( suppose

a < b ) touch at the point P . If circle A rotates x degree at P

anti-clockwisely . ( where 0 < x < 180 )

Find the area of the overlapping region of the 2 circles.

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- Apr 12th 2011, 01:07 AMyswongoverlapping area
2 circles A and B with radius a and b units respectively ( suppose

a < b ) touch at the point P . If circle A rotates x degree at P

anti-clockwisely . ( where 0 < x < 180 )

Find the area of the overlapping region of the 2 circles. - Apr 12th 2011, 03:51 AMearboth
1. Draw a sketch.

2. You'll get 2 sektors whose central angles are 2u and 2w respectively. You know that

$\displaystyle u+w=x$

The area of a segment is calculated by: Area of sector - area of right triangle.

The common side of the two triangles is calculated by:

$\displaystyle a \cdot \sin(u) = b \cdot \sin(w)~\implies~\dfrac ab = \dfrac{\sin(w)}{\sin(u)}$

Unfortunately LaTeX doesn't work anymore. Therefore you wasn't able to see this line:

**\tan(w)=\dfrac{a \cdot \sin(x)}{b+a\cdot \cos(x)}**

Re-written into the simple syntax it means:

**tan(w)=(a * sin(x))/(b+a*cos(x))**

3. Calculating the orange area $\displaystyle A_{orange}$:

$\displaystyle A_{orange} = \underbrace{\dfrac{2u}{360^\circ} \cdot \pi a^2}_{sector} - 2 \cdot \underbrace{\dfrac12 \cdot a^2 \cdot \cos(u) \cdot \sin(u)}_{\text{area of right triangle}}$

4. Calculating the green area $\displaystyle A_{green}$:

$\displaystyle A_{green} = \dfrac{2w}{360^\circ} \cdot \pi b^2 - 2 \cdot \dfrac12 \cdot b^2 \cdot \cos(w) \cdot \sin(w)$

5. Add both areas and try to simplify this term a little bit.

Maybe you can use the property: $\displaystyle \sin(u) \cdot \cos(u)=\dfrac12 \left(\sin(2u) \right)$ - Apr 18th 2011, 07:31 PMyswong
Thank you very much earboth! Your formula seems OK but can it be

expressed in terms of a , b and x only?

For special values of x :

(1) If x = 0 , the area should be 0.

(2) If x = 180 , the area should be a . a . pi sq.units. (i.e. area of circle A.)

(3) If x = 90 , what will be the area ?

Thanks again ! - Apr 18th 2011, 10:40 PMearboth
- Apr 23rd 2011, 05:03 AMyswong

Hi earboth , after revising some basic rules of trigonometry ,

I get the followings:

Let c be the distance between the centres of the 2 circles,

then c = √a*a+b*b-2ab cos(180-x)

= √a*a+b*b + 2ab cos x

Since a / c = sin w / sin (180 - x)

= sin w / sin x

Thus sin w = a * sin x / c

i.e. w = arc sin ( a * sin x / c )

Similarly u = arc sin (b * sin x / c )

Substituting into your formula ,

Orange area = u / 180 * π * a* a - a * a * sin (2u) / 2

Green area = w / 180 * π * b* b - b * b * sin (2w) / 2

Thus the total area can be expressed as a function of

a , b and x only . ( But I wonder how you get the expression :

tan(w)=(a * sin(x))/(b+a*cos(x))

For a special value of x being 90 ,

the overlapping area will be :

arc sin (b * sin 90 / c )* π * a* a / 180

+ arc sin (a * sin 90 /c )* π * b* b / 180

- the areas of the 2 triangles ( which = ab sq. units )

= arc sin ( b / c ) * π * a* a / 180 + arc sin (a /c )* π * b* b / 180

- ab

= arc sin ( b / √a*a+b*b )* π * a* a / 180

+ arc sin ( a / √a*a+b*b )* π * b* b / 180

- ab

Am I correct ? - May 3rd 2011, 12:43 AMyswong
Besides to find the overlapping area , we may also find the value

of c ( i.e. the distance between the centres of the 2 circles) given

by the formula : c = √a*a+b*b + 2ab cos x

( 1 ) If x = 0 , cos 0 = 1 , then c = √a*a+b*b + 2ab = a + b

(2) If x = 180 ,cos 180 = - 1, then c = √a*a+b*b - 2ab = b - a

If x is taken randomly from 0 to 180 , what will be the

expected value of c ?