Saturday, July 12, 2014

Geometry: Brining Circles Together (Illustrations done with Desmos)

Before I get started with today's blog post, I want to share an awesome online mathematical application.  All illustrations on this blog entry were done with the Desmos free on-line graphing calculator.  You can graph functions, parametric equations, table plots, and polar equations – complete with shading.  Many users have fun using Demos to draw interesting and fun graphics.  There is no monetary cost to signing an account.  You can check out Desmos at  

The Desmos Screen

The Two Circles
Suppose we have two circles, both having a radius of 1 and are touching at the origin of the Cartesian plane. (The origin is point (0,0).)

An equation that describes the two circles are:
Red Circle (on the left):  y^2 + (x + 1)^2 = 1
Blue Circle (on the right):  y^2 + (x – 1)^2 = 1

The two circles:  each of them are touching at the origin.

Simply put, the area of the two circles is 2*π. 

Moving the Red Circle In

What happens to the total area when I move the red circle to the right?  A portion of the circles will overlap, as illustrated below:

Moving the red circle to the right - note the overlapping ellipse shaded in purple.
If we want the total area, we have to take the overlapping ellipse (now shaded in purple) in mind.    One approach to calculating the area of the shape would be:

A = area of the two circles – area of the overlapping ellipse

A = 2 * π  - E

Where E = area of the ellipse

This ensures that there is not a portion of the shape, particularly the purple overlap that otherwise would get counted twice.   Let the variable p represent the distance that the red circle moves.  

The equations become:

Red Circle (on the left):  y^2 + (x + 1 - p)^2 = 1

Blue Circle (on the right):  y^2 + (x – 1)^2 = 1

In general, the area of any ellipse is:  A = π*a*b, where a and b are the length of the semi-major and semi-minor axes.  

The horizontal length of the ellipse is p.  Hence the semi-horizontal (semi-major or semi-minor depending on p) axis of the overlapping ellipse is p/2.  

Finding the semi-vertical axis will require finding the points where the two circles intersect.  Interestingly, the circles intersect when x=p/2.  The center point of the ellipse is (p/2, 0)
Using the equation for the red circle:

y^2 + (x + 1 – p)^2  = 1
y^2 + (p/2 + 1 – p)^2 = 1
y^2 + (1 – p/2)^2 = 1
y^2 = 1 – (1 – p/2)^2
y = ±√(1 – (1 – p/2)^2)
y = ±√(1 – (1 – p + p^2/4))
y = ±√(-p^2/4 + p)

To verify that we get the same answer, let’s substitute x=p/2 for the blue circle:

y^2 + (x – 1)^2 = 1
y^2 + (p/2 – 1)^2 = 1
y^2 = 1 – (p/2 – 1)^2
y = ±√(1 – (p/2 – 1)^2)
y = ±√(1 – (p^4/2 – p + 1))
y = ±√(-p^2/4 + p)

Hence the points where the circles interest are (p/2, ±√(-p^2/4 + p)).  The semi-vertical axis has a length of √(-p^2/4 + p).  The area of the overlapping ellipse, E, is:

E = π*p/2*√(-p^2/4 + p)

Overall, the area of the shape is:

A = 2*π - π*p/2*√(-p^2/4 + p)

Example:  p = 1/2
A = 2*π - π*(1/2)/2*√(-(1/2)^2/4 + 1/2)
A = 2*π – π/4*√(-1/16 + 1/2)
A = 2*π  - π/4 * √7/4 ≈ 5.76370


This blog is property of Edward Shore. 2014

1 comment:

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