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 www.desmos.com.
 |
| 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:
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| 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
Eddie
This blog is property
of Edward Shore. 2014
