Circles, Rotation & Quadrance

A Circle Is a Perpendicular Relationship Between Two Points

If a straight line projects linear extension, a circle is the locus of continuous perpendicular rotation of one point about another (G.10, G.15).

Center vs. Radius Point: Directional Asymmetry

Like a line, a circle is instantiated from two points—yet unlike a line, those two points are not interchangeable:

  • ( A B ) places the stationary center at \(A = (h, k)\) and sweeps through the radius point \(B\);

  • ( B A ) reverses their roles, centering at \(B\) and sweeping through \(A\).

Their shared radius squared is the quadrance \(Q(A, B)\):

\[Q(A, B) = (x_B - x_A)^2 + (y_B - y_A)^2 = r^2\]

and the algebraic locus of ( A B ) is purely quadratic in \((x, y)\) without a single square root in its defining polynomial:

\[(x - h)^2 + (y - k)^2 - Q(A, B) = 0\]

Quadratic Extensions & The Equidistant Poles

Whereas intersecting two lines stays inside the current rational field, intersecting a circle with a line or a second circle solves a quadratic equation—introducing quadratic surds (\(\sqrt{\Delta}\)) that unlock higher symmetries:

  1. First Compass Symmetry (`( A B ) ∩ ( B A )`): Intersecting the twin unit circles centered at \((0, 0)\) and \((1, 0)\) solves \(x = \frac{1}{2}\) and \(\left(\frac{1}{2}\right)^2 + y^2 = 1\), selecting the Equidistant Poles:

    \[C = \left(\frac{1}{2},\, \frac{\sqrt{3}}{2}\right), \qquad C' = \left(\frac{1}{2},\, -\frac{\sqrt{3}}{2}\right)\]

    In a single symmetric act, the Vesica Piscis generates the equilateral triangle (I.1 / G.17), perpendicularity (\([A B] \perp [C C']\)), bisection (\(M = (\tfrac{1}{2}, 0)\)), and the quadratic field \(\mathbb{Q}(\sqrt{3})\).

  2. Orthogonal Symmetry (`√2`): Sweeping a circle centered at midpoint \(M\) through \(A\) and \(B\) cuts the perpendicular pole line \([C C']\) at \(\left(\frac{1}{2}, \pm \frac{1}{2}\right)\), unfolding the square, right triangles, and \(\mathbb{Q}(\sqrt{2})\).

  3. Pentagonal & Golden Symmetry (`√5` and `φ`): Coupling the unit radius (\(1\)) with the bisected half-unit (\(\frac{1}{2}\)) along orthogonal axes produces hypotenuse quadrance \(Q = 1^2 + \left(\frac{1}{2}\right)^2 = \frac{5}{4}\), whose rotation onto the baseline yields \(\frac{1}{2} + \frac{\sqrt{5}}{2} = \phi\) (II.11, IV.10–11).