Associated Topics || Dr. Math Home || Search Dr. Math

### Finding the Area of Overlapping Circles

```Date: 11/08/2004 at 21:31:10
From: Nick
Subject: Area of Overlaping Circles

Given two circles:

circle O: (x - 1)^2 + (y - 2)^2 = 25
circle P: (x - 4)^2 + y^2 = 16

What is the area of the overlapping part of the circles?

I couldn't find anything on how to do this kind of problem, and I'm stuck.

```

```
Date: 11/12/2004 at 11:18:59
From: Doctor Cristian
Subject: Re: Area of Overlaping Circles

Hi Nick, and thanks for asking Dr. Math!

I can think of no other way than the straight-forward one, but when
I try to put it all together for the general case, I found that
there's no "nice" answer, so I can't give you a formula to help with
other similar cases.  So bear with me, and we'll be getting to the

First, note that the relative position of the two circles (and thus
the area of their intersection) doesn't change upon moving and
rotating the coordinate system.  So we can say that one circle is
centered in the origin, and the other one is centered at (d, 0),
where d is the original distance between the two centers.  This is
purely for easing our work, since we could, in theory, work with the
given data.

The actual value for d, in your case, is

d = sqrt( (x1-x2)^2 + (y1-y2)^2 )
= sqrt( (1-4)^2 + (2-0)^2 )
= sqrt( 9 + 4 )
= sqrt(13)

So now that our picture looks a little better (you can look at

http://mathforum.org/dr.math/gifs/nick.11.12.04.jpg

to see a sketch of it), we should find the coordinates of the two
intersection points.  That can be done by solving the system of the
equations of the circles:

x^2 + y^2 = r1^2

(x-d)^2 + y^2 = r2^2

Substracting the two, we get

-2dx + d^2 = r2^2 - r1^2, or

x = (r1^2 + d^2 - r2^2) / (2d), so

y^2 = r1^2 - x^2

= 2 r1^2 d^2 + 2 r2^2 d^2 + 2 r1^2 r2^2 - r1^4 - r2^4 - d^4
---------------------------------------------------------
4 d^2

which gives us the two solutions (one positive and one negative):

yp = sqrt( ... ), and

yn = -sqrt( ... )

x = (25 + 13 - 16) / (2sqrt(13)) = 11 sqrt(13) / 13

yp = sqrt( 25 - (121/13) ) = sqrt(204/13) = 2 sqrt(51/13)

= 2 sqrt(663) / 13

yn = -yp = -2 sqrt(663) / 13

Now we can find the angles that these chords (it is only one segment,
but it is a *different* chord in each circle) subtend.

angle = 2 arcsin( (length/2) / radius ), where length is

length = yp - yn

In the first circle,

a1 = 2 arcsin ( (2 sqrt(663) / 13) / 5 )

= 2 arcsin ( 2 sqrt(663) / 65 ), which approximates to

~ 2 * 0.9145 = 1.829 (radians),

which is about 105 degrees--this can be checked in the picture.

In the second circle,

a2 = 2 arcsin ( (2 sqrt(663) / 13) / 4 )

= 2 arcsin ( sqrt(663) / 26 ), which approximates to

~ 2 * 1.4317 = 2.8634 (radians),

which is about 164 degrees--this too can be checked in the pic.

And now the two segments of the circles come into play.  You can
review what they are and a few formulas about them at

Circle Formulas
http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment

The area is simply the area of the sector of circle minus the area
of the triangle.  Keep in mind though that the segment of circle that
lies closer to the center of one of the two circles actually belongs
to the *other* circle!

The area of the segment belonging to the first circle is

A_seg_1 = r1^2 a1 / 2 = 25 arcsin ( 2 sqrt(663) / 65 )

The area of the corresponding triangle is

A_tri_1 = x * yp = (11 / sqrt(13)) * (2 sqrt(51/13))
= 22 sqrt(51) / 13

So the area of the righternmost half of the intersection is

A_sect_1 = A_seg_1 - A_tri_1

= 25 arcsin ( 2 sqrt(663) / 65 ) - 22 sqrt(51) / 13

Similarly, the other three areas are

A_seg_2 = r2^2 a2 / 2 = 16 arcsin ( sqrt(663) / 26 )

A_tri_2 = (d - x) * yp = (2 / sqrt(13)) * (2 sqrt(51/13))

= 4 sqrt(51) / 13

A_sect_2 = A_seg_2 - A_tri_2

= 16 arcsin ( sqrt(663) / 26 ) - 4 sqrt(51) / 13

And, finally, we've found the answer of

A_intersection

= A_sect_1 + A_sect_2

=    25 arcsin ( 2 sqrt(663) / 65 ) - 22 sqrt(51) / 13
+ 16 arcsin ( sqrt(663) / 26 ) - 4 sqrt(51) / 13
~ 25 * 0.9145 + 16 * 1.4317 - 2 * 7.1414 = 31.487

awkward-looking process.

- Doctor Cristian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Conic Sections/Circles

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search