Rational Trigonometry: Spread in a circle Part 2
Spread measurement outside a circle
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By symmetry, the spread at vertex B can be seen to equal the spread at several other vertices in the above figure, for example the spread at vertex C:
S(C) = x²/h² = x² / (x² + y²) = x²/(x² + x (1 – x)) = x
And, even more directly:
S(B) = h²/1² = (x² + y²) = (x² + x (1 – x)) = x
(In a unit circle therefore, the quadrance of a chord equals the opposite subtended spread, and is given by the chord’s projection on a diameter on which it sits)
We now introduce a new form of quadrance, based on a difference rather than a sum of squares – and measure its associated spread value, so-called red quadrance (and spread):
QR(1,2) = (x2 – x1)² – (y2 – y1)².
From similarity in the figure above we have:
ED/EB (in DEB) = y’/ (1 + x’)
= EA/ED (in AED) = x’/y’,
y’/ (1 + x’) = x’/y’ ⇒ y’² = (x’)(1 + x’)
in terms of the definition of red quadrance therefore,
QR(AD) = x’² – y’² = x’² – (x’)(1 + x’) = -x’
Note that the vertex at D with spread S(DA,DB) is formed by reflecting AC (perpendicular to DB) through the vertical line at A, making lines DA and DB symmetrical with respect to lines of slope +1 and -1 through D as may be verified from the figure above (AD is rotated as far from the vertical – and in an opposite sense – as BD is rotated from the horizontal). This condition of being reflected in lines at ’45 degrees’ to the usual coordinate axes defines perpendicularity in the geometry of red quadrances and spreads.
Specifically, if one line passes through a point in direction (a,b) from the origin, and another passes through a point lying in direction (b’,a’) from that origin, the lines are red perpendicular since their red dot product is 0. That is, for any scalars k1 and k2 (with a=a’ and b=b’)
k1(a, b) dotR k2(b’, a’) = k1k2(ab’ – (ba’)) = 0 ⇔ k1(a,b) ⊥ k2,(b’,a’)
This contrasts with the usual Euclidean dot product (which henceforth we identify by the prefix blue) that vectors in directions (a,b) and (b’,-a’) through the origin are blue perpendicular since, for any scalars k1 and k2
k1(a, b) dotB k2(b’, -a’) = k1k2(ab’ + b(-a’)) = 0 ⇔ k1(a,b) ⊥ k2,(b’,-a’)
Thus in triangle DAB vertex D is a red perpendicular, so QR(AD)/QR(AB) defines an equivalent spread ratio at (opposite) vertex B. Being parallel to the x direction, the (red) hypotenuse quadrance of diameter AB remains a unit and therefore
SR(B) = QR(AD)/1 = –x’
Note that the sign of this red spread and quadrance corresponds to the displacement of segment AE to the left of the ‘0’ position used in the measurement of blue spread. This is not accidental. Allowing for sign, the red spread of vertex B corresponds to the x coordinate of the exterior similar triangle formed.
Finally we note the locus of point D, as segment AC is moved on auxiliary circle, (x – ½)² + y² = (½)² is the rectangular hyperbola (x – ½)² – y² = (½)², with a common centre at (½,0), touching on common ‘diameter’ AB. Accordingly we henceforth identify the locus of red perpendicularity on this diameter as a red circle and our previous circle with the prefix blue, putting both objects on equal footing. The noted equal but opposite rotations of lines passing through fixed points A and B always meet (red) perpendicularly on this curve, just as equal rotations of lines passing through fixed points will meet (blue) perpendicularly on a classic circle.