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Topic: Nbic Circles: Chord distance
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Doctor Nisith Bairagi

Posts: 27
From: Uttarpara, West Bengal, India
Registered: 3/2/13
Nbic Circles: Chord distance
Posted: Apr 10, 2013 10:26 AM
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From: Doctor Nisith Bairagi
Date: April 10, 2013
The generalized form of chord distance of Nbic (viz., Euclidean, Hyperbolic, Nbic) circles has been presented. Reference may be made to the Topic <CLASSIFICATION OF CIRCLES:A MAIDEN CONCEPT> posted on March 19, 2013, and on March 20, 2013 in (geometry.research.indepedent).
Distance (d) between any two points A(X1, Y1) and B(X2, Y2) is given by:
d = AB = sqrt[(X2-X1)^2 + (Y2-Y1)^2]

(A) For Euclidean circle (plane circle), C<0>:

X^2 + Y^2 = 1, (X, Y) = (cos x, sin x), [where, x is angle in rad (or, in deg)],
the chord distance d turns out to be:
d = sqrt[2].sqrt[1-(cos x2.cos x1 + sin x2.sin x1)] = 2 sin[(x2-x1)/2]
With this, we get:
Between x1 = 0, and, x2 = PI/2, d = sqrt[2]
and, x1 = 0, x2 = PI, d = 2 (diameter).

(B) For Hyperbolic circle (plane circle), C<H>:

X^2 + Y^2 = 1, (X, Y) = [cosh x/sqrt(cosh 2x), sinh x/sqrt(cosh 2x)]
the chord distance d turns out to be:
d = sqrt([2].sqrt[1-(cosh x2.cosh x1 + sinh x2.sinh x1)/(cosh 2x2.cosh 2x1)]
= sqrt([2].sqrt[{cosh(x2 + x1)}/(cosh 2x2.cosh 2x1)]

(C) For all other Nbic circles, C<N>:
[viz., C<1>, C<2e>, C<2f>, C<3e>, C<3f>], satisfying, X^2 + Y^2 = 1,

The distance formula (d) for Nbic circles C<N, is written using parameters (k and m):

C<1>: (k = <1>, m = 1), (X, Y) = [cosN(x,x), sinN(x, x)]/sqrt(cosh 2x)
C<2e>: (k = <2e>, m = 1), (X, Y) = [cosN<2e>(x, x), sinN<2e>(x, x)]/sqrt(cosh 2x)
C<2f>: (k = <2f>, m = 2), (X, Y) = [cosN<2f>(x, x), sinN<2f>(x, x)]/cosh 2x
C<3e>: (k = <3e>, m = 2), (X, Y) = [cosN<3e>(x, x), sinN<3e>(x, x)]/cosh 2x
C<3f>: (k = <3f>, m = 3), (X, Y) = [cosN<3f>(x, x), sinN<3f>(x, x)]/(cosh 2x)^(3/2):

d = sqrt([2].sqrt[1-(cosN<k> x2.cosN<k> x1 + sinN<k> x2.sinN<k> x1)/
(cosh 2x2.cosh 2x1)^m]

Please comment on this topic for betterment.

Thanks to ALL

Doctor Nisith Bairagi

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