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Area of Union of Two Circles


Date: 6/10/96 at 10:26:43
From: West Craig
Subject: Area of union of two circles

Hi,

I understand from Doctor Bob's Accessing The Internet By E-Mail 
that you guys can help with maths problems. Well here goes...

(1) Take a circular field of radius R 
(2) Plant a tree at some point on the circumference of the circular 
    field 
(3) Tie a rope to the tree and attach the other end to a passing goat 
(4) If the effective length of the rope is L, and the goat can eat 
    exactly half of the grass in the field, express L in terms of R

Any solution would be greatly appreciated

Cheers,
Craig West


Date: 6/10/96 at 11:42:51
From: Doctor Anthony
Subject: Re: Area of union of two circles

I will slightly change the notation, to r = radius of given circle, 
and R = radius of circle defined by L the length of rope.
Since one cannot draw a decent diagram with ascii, I suggest you
draw the following diagram on a piece of paper and refer to it while
I go through the working.  

Draw a circle with suitable radius r. Now take a point C on the 
circumference and with a slightly larger radius R draw an arc of a 
circle to cut the first circle in points A and B. Join AC and BC. 
Let O be the centre of the first circle of radius r. Let angle OCA = x 
(radians). This will also be equal to angle OCB. The area we require 
is made up of a sector of a circle radius R with angle 2x at the 
centre, C, of this circle, plus two small segments of the first circle 
of radius r cut off by the chords AC and BC.

The area of the sector of circle R is (1/2)R^2*2x = R^2*x

Area of two segments = 2[(1/2)r^2(pi-2x) - (1/2)r^2sin(pi-2x)]
                     = r^2[pi - 2x - sin(2x)]

We also have R = 2rcos(x)   so R^2*x = 4r^2*x*cos^2(x)

We add the two elements of area and equate to (1/2)pi*r^2

4r^2*x*cos^2(x) + r^2[pi-2x-sin(2x)] = (1/2)pi*r^2   divide out r^2
    4x*cos^2(x) + pi - 2x - sin(2x)  = (1/2)pi
4x*cos^2(x) + (1/2)pi - 2x - sin(2x) = 0

We must solve this for x and we can then find R/r from R/r = 2cos(x)

Newton-Raphson is a suitable method for solving this equation, using
a starting value for x at about 0.7 radians

The solution I get is x = 0.95284786466 and from this
                 cos(x) = 0.579364236509

and so finally  R/r = 2cos(x) = 1.15872847

In terms of the notation you used this is 

  L = 1.1587*radius of given circle. 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


    
Associated Topics:
High School Calculus
High School Conic Sections/Circles
High School Geometry

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