8P-s-P1: Cayley-Bacharach Point

The Cayley-Bacharach point in an Octangle is defined as follows:
Let P1,...,P8 be eight distinct points in the plane, no three of which collinear, and no six of them on a conic. There exists a unique ninth point P9 such that every cubic curve through P1,...,P8 also contains P9.

In next picture an example is given of 4 vertices of a Reference Quadrangle combined with the centroids of the 4 component triangles (together 8 points) producing a Cayley Bacharach point common for all cubics through this 8 points.
 
8P s P1 Cayley Bacharach Point

Many properties can be derived from the Cayley Bacharach Theorem, especially when a cubic is degenerated, for example in a conic and two lines, etc.

See also QFG-messages #732, #733, #736, #737, #1476, #1491, #1517, #1696, #1698-1701, #1733.

Calculation of coordinates:
In http://arxiv.org/pdf/1405.6438v2.pdf (Cayley-Bacharach Formulas by Qingchun Ren, Jürgen Richter-Gebert and Bernd Sturmfels) is given a method for calculating this point.

Let P1 = (x1:y1:z1), P2 = (x2:y2:z2), etc.
Let C(P1,P2,P3,P4,P5,P6) = Determinant of

8P s P1 CB Point Determinat 1
 
Let D(P1;P2,P3,P4,P5,P6,P7,P8) = Determinant of

8P s P1 CB Point Determinat 2

Cx = C(P1,P4,P5,P6,P7,P8),

Cy = C(P2,P4,P5,P6,P7,P8),
Cz = C(P3,P4,P5,P6,P7,P8),
Dx = D(P1;P2,P3,P4,P5,P6,P7,P8),
Dy = D(P2;P3,P1,P4,P5,P6,P7,P8),
Dz = D(P3;P1,P2,P4,P5,P6,P7,P8).

The Cayley-Bacharach point is given by the formula:

                                     P9 = CxDyDz · P1 + DxCyDz · P2 + DxDyCz · P3


Correspondence with EQF:
Eckart Schmidt wrote next properties of the Cayley Bacharach Point (CB-point) in relationship to Quadrangles/Quadrigons at Ref-34, QFG#2640:
(1) Consider the vertices of a quadrangle and their complements wrt the remaining triangle.
The CB-point of these 8 points is QA-P1.
(2) Consider the vertices of a quadrangle and their anticomplements wrt the remaining triangle.
The CB-point of these 8 points is a new QA-point
... on a QA-circumscribed conic through QA-P10 and QA-P5
... and a parallel to QA-P1.QA-P16 through QA-P5,
... with tangent through QA-P16.
(3) Consider the vertices of a quadrangle and the in-/ex-centers of the Miquel triangle QA-Tr2.
The CB-point of these 8 points
... is the 3rd intersection of QA-Cu1 and Q.QA-P41
... with Q = intersection of QA-Cu1 and its asymptote (see EQF).
(4) Consider the vertices of a quadrangle and their isogonal conjugates wrt the Miquel triangle,
The CB-point of these 8 points is the tangential of Q wrt QA-Cu1.
(5) Consider the vertices of a quadrangle, its mid-DT and QA-P10.
The CB-point of these 8 points is the 2nd intersection of QA-P16.QA-P20 and the QA-circumconic through QA-P20.
(6) Consider the vertices of a quadrigon and the 4th parallelogram points of three consecutive vertices.
The CB-point of these 8 points is the reflection of QG-P1 in QG-P15.
(The QA-triangle of these CB-points is homothetic to QA-Tr1 wrt QA-P1 and factor -3.)
(7) Consider the vertices of QA-Tr1 and QA-Tr2 on QA-Cu1
... with Q and QA-P3: CB-point = QA-Tf2(res(QA-P4,QA-P41)) = QA-P41,
... with Q and QA-P4: CB-point = QA-Tf2(res(QA-P3,QA-P41)),
... with QA-P3 and QA-P4: CB-point = QA-Tf2(res(Q,QA-P41),
... with QA-P4 and QA-P41: CB-point = QA-Tf2(res(Q,QA-P3)).
Or: If you take the vertices of QA-Tr1 and QA-Tr2 and two of the points Q, QA-P3, QA-P4, QA-P41,
you get the CB-point as QA-Tf2-image of the 3rd intersection of QA-Cu1 and the line through the other two points.
(8) Consider the vertices of a quadrangle and Q, QA-P3, QA-P4, QA-P41.
The CB-point of these 8 points is res(Q,QA-Tf2(res(QA-P3,QA-P41))), where res(X, Y) = 3rd intersection of QA-Cu1 and XY.

Other Properties:
Let P1P2P3P4 be a quadrangle and when P5P6P7 is its diagonal triangle and P8 is some random point, then P9 = QA-Tf2(P8). See Ref-34, Seiichi Kirikami, QFG#1698.