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#### Rational Trigonometry: ** Spread in a circle ****Part 2**

*Spread* measurement *outside* a circle

Click for a larger image

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):

Q_{R}(1,2) = (x_{2} – x_{1})² – (y_{2} – y_{1})².

From similarity in the figure above we have:

** ED**/**EB** (in **DEB**) = y’/ (1 + x’)

= **EA**/**ED** (in **AED**) = x’/y’,

therefore

y’/ (1 + x’) = x’/y’ ⇒ y’² = (x’)(1 + x’)

in terms of the definition of red quadrance therefore,** **

Q_{R}(**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 k_{1} and k_{2}** (with a=a’ and b=b’)**

k_{1}(**a, b**) dot_{R} k_{2}(**b’, a’**) = k_{1}k_{2}(ab’ **–** (ba’)) = 0 ⇔ k_{1}(**a,b**) **⊥** k_{2},(**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 k_{1} and k_{2}

k_{1}(**a, b**) dot_{B} k_{2}(**b’, -a’**) = k_{1}k_{2}(ab’ **+** b(-a’)) = 0 ⇔ k_{1}(**a,b**) **⊥** k_{2},(**b’,-a’**)** **

Thus in triangle **DAB** vertex **D** is a red perpendicular, so Q_{R}(**AD**)/Q_{R}(**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** **

S_{R}(**B**) = Q_{R}(**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.

**Start**

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