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Simplifying Circle Formulas from the Dr. Math FAQ

Date: 04/30/2004 at 14:47:14
From: Jim
Subject: faq.circle.segment.html

In your FAQ on circle formulas, in the sections where the other five
values are derived from any two known values, could you write each
formula in terms of only the two known values, instead of using the
intermediate steps?

For example, in Case 10 I know c and d, so can all the formulas be
expressed simply in terms of c and d?  Can the trig function be
represented more concisely?  It's even more of an issue for K, where I
would need to take the sin of an arcsin to get to the answer.



Date: 04/30/2004 at 15:54:41
From: Doctor Peterson
Subject: Re: faq.circle.segment.html

Hi, Jim.

Certainly you can combine formulas to get an individual result 
directly from the givens, if you don't need the other values, and 
sometimes that will give neat results.  In Case 10, you are told

   r     = sqrt(c^2 + 4d^2)/2
   h     = r - d
   theta = 2 arcsin(c/[2r])
   s     = r theta
   K     = r^2[theta - sin(theta)]/2

As you suggest, combining to get s directly gives the unwieldy

  s = sqrt(c^2 + 4d^2) * arcsin(c/sqrt(c^2 + 4d^2))

To simplify the arcsin, look at this triangle:

                      +
  sqrt(c^2 + 4d^2) /  |
                /     |c
             /        |
          +-----------+

What is the other side?

  sqrt[c^2 + 4d^2 - c^2] = sqrt(4d^2) = 2d

So the tangent of the angle is c/(2d), and our formula simplifies to

  s = sqrt(c^2 + 4d^2) * arctan(c/(2d))

I think that helps a bit.

Now, for K, we are taking the sine of twice the angle whose sine is 
c/sqrt(c^2 + 4d^2).  We can use the double-angle identity
 
  sin(2x) = 2 sin(x) cos(x) = 2 sin(x) sqrt(1 - sin^2(x))

to get, for the sine,

  2 c/sqrt(c^2 + 4d^2) sqrt[1 - c^2/(c^2 + 4d^2)]

which simplifies nicely to

     4cd
  ----------
  c^2 + 4d^2

(check that and make sure I'm right!), so the formula becomes

  K = (c^2 + 4d^2)/4 [2 arctan(c/(2d)) - 4cd/(c^2 + 4d^2)]/2

    = (c^2 + 4d^2)/2 arctan(c/(2d)) - cd/2

Still not simple, but a lot nicer that it would have been without the 
simplifications.

Now, you can also get these formulas directly by going back to the 
geometry.  You have this figure:

                      ooooooooo
                oooooo....|....oooooo
             ooo..........|..........ooo
           +--------------+--------------+
         oo \             |    c/2      / oo
        o     \           |           /     o
       o        \        d|         /        o
      o           \       |       /           o
     o              \     |     /              o
     o                \   |t/2/                o
    o                   \ | /                   o
    o                     +                     o
    o                     |                     o
     o                    |                    o
     o                    |                    o
      o                   |                   o
       o                  |                  o
        o                 |                 o
         oo               |               oo
           oo             |             oo
             ooo          |          ooo
                oooooo    |    oooooo
                      ooooooooo

The area will be the area of the sector minus the area of the 
isosceles triangle.  The latter is just cd/2 (sounds familiar!), and 
the former is r^2 theta/2, where theta/2 (shown as t/2 in the figure) 
has tangent (c/2)/d.  So

 K = r^2/2 arctan(c/(2d)) - cd/2

which is just what we had above, once we replace r with its value.

I wouldn't mind seeing more of these formulas in the FAQ, since it 
takes a bit of work to get them one way or the other.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 04/30/2004 at 16:05:41
From: Jim
Subject: Thank you (faq.circle.segment.html)

Thanks!  That was fast, thorough, and it even included an ASCII 
graphic!

- Jim
Associated Topics:
High School Conic Sections/Circles
High School Trigonometry

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