Editing Ellipse (section)

== Inscribed angles and three-point form ==
=== Circles ===
[[File:Inscribe-a-c.svg|thumb|Circle: inscribed angle theorem]]
A circle with equation <math>\left(x - x_\circ\right)^2 + \left(y - y_\circ\right)^2 = r^2</math> is uniquely determined by three points <math>\left(x_1, y_1\right),\; \left(x_2,\,y_2\right),\; \left(x_3,\, y_3\right)</math> not on a line. A simple way to determine the parameters <math>x_\circ,y_\circ,r</math> uses the ''[[inscribed angle theorem]]'' for circles:
: For four points <math>P_i = \left(x_i,\, y_i\right),\ i = 1,\, 2,\, 3,\, 4,\,</math> (see diagram) the following statement is true:
: The four points are on a circle if and only if the angles at <math>P_3</math> and <math>P_4</math> are equal.

Usually one measures inscribed angles by a degree or radian ''θ'', but here the following measurement is more convenient:
: In order to measure the angle between two lines with equations <math>y = m_1x + d_1,\ y = m_2x + d_2,\ m_1 \ne m_2,</math>  one uses the quotient: <math display="block">\frac{1 + m_1 m_2}{m_2 - m_1} = \cot\theta\ .</math>

====Inscribed angle theorem for circles====
For four points  <math>P_i = \left(x_i,\, y_i\right),\ i = 1,\, 2,\, 3,\, 4,\,</math> no three of them on a line, we have the following (see diagram):

: The four points are on a circle, if and only if the angles at <math>P_3</math> and <math>P_4</math> are equal. In terms of the angle measurement above, this means: <math display="block">
\frac{(x_4 - x_1)(x_4 - x_2) + (y_4 - y_1)(y_4 - y_2)}
     {(y_4 - y_1)(x_4 - x_2) - (y_4 - y_2)(x_4 - x_1)} =
\frac{(x_3 - x_1)(x_3 - x_2) + (y_3 - y_1)(y_3 - y_2)}
     {(y_3 - y_1)(x_3 - x_2) - (y_3 - y_2)(x_3 - x_1)}.
</math>

At first the measure is available only for chords not parallel to the y-axis, but the final formula works for any chord.

====Three-point form of circle equation====

: As a consequence, one obtains an equation for the circle determined by three non-collinear points <math>P_i = \left(x_i,\, y_i\right)</math>: <math display="block">
  \frac{({\color{red}x} - x_1)({\color{red}x} - x_2) + ({\color{red}y} - y_1)({\color{red}y} - y_2)}
       {({\color{red}y} - y_1)({\color{red}x} - x_2) - ({\color{red}y} - y_2)({\color{red}x} - x_1)} =
  \frac{(x_3 - x_1)(x_3 - x_2) + (y_3 - y_1)(y_3 - y_2)}
       {(y_3 - y_1)(x_3 - x_2) - (y_3 - y_2)(x_3 - x_1)}.
</math>

For example, for <math>P_1 = (2,\, 0),\; P_2 = (0,\, 1),\; P_3 = (0,\,0)</math> the three-point equation is:
: <math>\frac{(x - 2)x + y(y - 1)}{yx - (y - 1)(x - 2)} = 0</math>, which can be rearranged to <math>(x - 1)^2 + \left(y - \tfrac{1}{2}\right)^2 = \tfrac{5}{4}\ .</math>

Using vectors, [[dot product]]s and [[determinant]]s this formula can be arranged more clearly, letting <math>\vec x = (x,\, y)</math>:
<math display="block">
  \frac{\left({\color{red}\vec x} - \vec x_1\right) \cdot \left({\color{red}\vec x} - \vec x_2\right)}
       {\det\left({\color{red}\vec x} - \vec x_1,{\color{red}\vec x} - \vec x_2\right)} =
  \frac{\left(\vec x_3 - \vec x_1\right) \cdot \left(\vec x_3 - \vec x_2\right)}
       {\det\left(\vec x_3 - \vec x_1, \vec x_3 - \vec x_2\right)}.
</math>

The center of the circle <math>\left(x_\circ,\, y_\circ\right)</math> satisfies:
<math display="block">\begin{bmatrix}
    1 & \dfrac{y_1 - y_2}{x_1 - x_2} \\[2ex]
    \dfrac{x_1 - x_3}{y_1 - y_3} & 1
  \end{bmatrix}
  \begin{bmatrix} x_\circ \\[1ex] y_\circ \end{bmatrix}
  =
  \begin{bmatrix}
    \dfrac{x_1^2 - x_2^2 + y_1^2 - y_2^2}{2(x_1 - x_2)} \\[2ex]
    \dfrac{y_1^2 - y_3^2 + x_1^2 - x_3^2}{2(y_1 - y_3)}
  \end{bmatrix}.
</math>

The radius is the distance between any of the three points and the center.
<math display="block">
  r = \sqrt{\left(x_1 - x_\circ\right)^2 + \left(y_1 - y_\circ\right)^2}
    = \sqrt{\left(x_2 - x_\circ\right)^2 + \left(y_2 - y_\circ\right)^2}
    = \sqrt{\left(x_3 - x_\circ\right)^2 + \left(y_3 - y_\circ\right)^2}.
</math>

=== Ellipses ===
This section considers the family of ellipses defined by equations <math>\tfrac{\left(x - x_\circ\right)^2}{a^2} +  \tfrac{\left(y - y_\circ\right)^2}{b^2} = 1</math> with a ''fixed'' eccentricity <math>e</math>. It is convenient to use the parameter:
<math display="block">{\color{blue}q} = \frac{a^2}{b^2} = \frac{1}{1 - e^2},</math>

and to write the ellipse equation as:
<math display="block">\left(x - x_\circ\right)^2 + {\color{blue}q}\, \left(y - y_\circ\right)^2 = a^2,</math>

where ''q'' is fixed and <math>x_\circ,\, y_\circ,\, a</math> vary over the real numbers. (Such ellipses have their axes parallel to the coordinate axes: if <math>q < 1</math>, the major axis is parallel to the ''x''-axis; if <math>q > 1</math>, it is parallel to the ''y''-axis.)

[[File:Inscribe-a-e.svg|thumb|Inscribed angle theorem for an ellipse]]
Like a circle, such an ellipse is determined by three points not on a line.

For this family of ellipses, one introduces the following [[q-analog]] angle measure, which is ''not'' a function of the usual angle measure ''θ'':<ref>[http://www.mathematik.tu-darmstadt.de/~ehartmann/circlegeom.pdf E. Hartmann: Lecture Note '<nowiki/>'''Planar Circle Geometries'''', an Introduction to Möbius-, Laguerre- and Minkowski Planes, p. 55]</ref><ref>W. Benz, ''Vorlesungen über Geomerie der Algebren'', [[Springer Science+Business Media|Springer]] (1973)</ref>
: In order to measure an angle between two lines with equations <math>y = m_1x + d_1,\ y = m_2x + d_2,\ m_1 \ne m_2</math> one uses the quotient: <math display="block">\frac{1 + {\color{blue}q}\; m_1 m_2}{m_2 - m_1}\ .</math>

====Inscribed angle theorem for ellipses====

: Given four points <math>P_i = \left(x_i,\, y_i\right),\ i = 1,\, 2,\, 3,\, 4</math>, no three of them on a line (see diagram).
: The four points are on an ellipse with equation <math>(x - x_\circ)^2 + {\color{blue}q}\, (y - y_\circ)^2 = a^2</math> if and only if the angles at <math>P_3</math> and <math>P_4</math> are equal in the sense of the measurement above—that is, if <math display="block">
\frac{(x_4 - x_1)(x_4 - x_2) + {\color{blue}q}\;(y_4 - y_1)(y_4 - y_2)}
     {(y_4 - y_1)(x_4 - x_2) - (y_4 - y_2)(x_4 - x_1)} =
\frac{(x_3 - x_1)(x_3 - x_2) + {\color{blue}q}\;(y_3 - y_1)(y_3 - y_2)}
     {(y_3 - y_1)(x_3 - x_2) - (y_3 - y_2)(x_3 - x_1)}\ .
</math>

At first the measure is available only for chords which are not parallel to the y-axis. But the final formula works for any chord. The proof follows from a straightforward calculation. For the direction of proof given that the points are on an ellipse, one can assume that the center of the ellipse is the origin.

====Three-point form of ellipse equation====

: A consequence, one obtains an equation for the ellipse determined by three non-collinear points <math>P_i = \left(x_i,\, y_i\right)</math>: <math display="block">
\frac{({\color{red}x} - x_1)({\color{red}x} - x_2) + {\color{blue}q}\;({\color{red}y} - y_1)({\color{red}y} - y_2)}
     {({\color{red}y} - y_1)({\color{red}x} - x_2) - ({\color{red}y} - y_2)({\color{red}x} - x_1)} =
\frac{(x_3 - x_1)(x_3 - x_2) + {\color{blue}q}\;(y_3 - y_1)(y_3 - y_2)}
     {(y_3 - y_1)(x_3 - x_2) - (y_3 - y_2)(x_3 - x_1)}\ .
</math>

For example, for <math>P_1 = (2,\, 0),\; P_2 = (0,\,1),\; P_3 = (0,\, 0)</math> and <math>q = 4</math> one obtains the three-point form
: <math>\frac{(x - 2)x + 4y(y - 1)}{yx - (y - 1)(x - 2)} = 0</math> and after conversion <math>\frac{(x - 1)^2}{2} + \frac{\left(y - \frac{1}{2}\right)^2}{\frac{1}{2}} = 1.</math>

Analogously to the circle case, the equation can be written more clearly using vectors:
<math display="block">
    \frac{\left({\color{red}\vec x} - \vec x_1\right)*\left({\color{red}\vec x} - \vec x_2\right)}
         {\det\left({\color{red}\vec x} - \vec x_1,{\color{red}\vec x} - \vec x_2\right)}
  = \frac{\left(\vec x_3 - \vec x_1\right)*\left(\vec x_3 - \vec x_2\right)}
         {\det\left(\vec x_3 - \vec x_1, \vec x_3 - \vec x_2\right)},
</math>

where <math>*</math> is the modified [[dot product]] <math>\vec u*\vec v = u_x v_x + {\color{blue}q}\,u_y v_y.</math>