Analogously is defined the conjugacy of two lines: Two lines {x,y} are said to be

The symmetry of the relation is due to the reciprocity of poles-polars: if Y is on the polar of X, then also X is on the polar of Y. If X' is the projection of X on its polar p

From this follows that all circles with diameter XY are orthogonal to (c) and being also orthogonal to line (e) belong to the bundle of circles (II) which is orthogonal to the circle bundle (I) generated by {c, e}.

The kind of the circle-bundle (II) depends on the position of line (e) relative to the circle c.

[1] If line (e) does not intersect the circle c, then bundle (II) is of intersecting type. All the circles of (II) pass through two fixed points {E',E''} on the line orthogonal to (e) through the center O of the circle. This line is the common radical axis of all circles of the bundle (II). Points {E', E''} are harmonic conjugate with respect to {O, E}, E being the pole of (e).

In this case the involution F has no fixed points on line (e).

[2] If line (e) intersects the circle (c), then bundle (II) is of non-intersecting type with limit points the intersection points {X

In this case the involution F has two fixed points on line (e) which are precisely {X

The definition of line-conjugacy can be transferred also verbatim to this general case.

The above basic figure shows two points {X,Y} on line (e) conjugate with respect to a conic c. Point E is the pole of (e) and lines {Y'Y'', X'X''} are respectively the polars of {X, Y}. By Pascal's theorem the opposite sides of quadrangle X'Y'X''Y'' intersect at two points {Z, Z'} which are on (e) and are also related by the homographic relation Z'=F(Z). Besides {Z,Z'} are harmonic conjugate with respect to {X,Y}.

The previous statements on the circle bundle (II) of circles with diameters XY transfer almost unchanged to this case. The general figure can be transformed by an appropriate projectivity to a circle (c') and a line (e'), the homography relation being preserved, we obtain also a map of the bundle (II) here to the corresponding bundle of {c', e'}. We conclude that the circle bundle in the present general case is of intersecting type or non-intersecting type in dependence of the position of the line e to the conic c:

[1] If the line (e) does not intersect the conic then the bundle (II) of circles with diameters XY is of intersecting type. All circles pass through two fixed points {E',E''}, which can be explicitly constructed from the data.

The involutive homgoraphy Y = F(X) has no fixed points in this case.

[2] If the line (e) intesects the conic (c) at two points {X

[1] Because {Z,Z'} are harmonic conjugate to {X,Y} the center of the circle c

[2] The previous position of the center of the circle is thus the pole Z

[3] Another interesting circle of the circle-bundle (II) results when X obtains the position of Z

[4] The previous arguments show that Z

[5] Points {E', E''} can though be constructed also geometrically by noticing the relation of pairs of points {Z,Z'} and {X,Y} in the figure of section 2. Namely one pair is defined by the diagonals of a quadrangle inscribed in the conic whereas the other is defined by the common points of pairs of opposite sides. Applying this remark to the pair consisting of Z

[6] From what has been said {E

To study this involution (of conjugacy of diameters in a central conic) we identify the line at infinity with the pencil O* of lines through O. Then our involution becomes an involution of this pencil and can be studied by taking the traces X, Y , ... of lines through O on a fixed line e. This new involution associates to each point X of line (e) a point Y such that {OX, OY} are conjugate diameters of the conic. There is again a naturally associated pair of conjugate points {Z,Z'} to {X,Y} comming from the diameters parallel to the sides of the parallelogram with diagonals along {OX, OY}. By analogous arguments as in the previous section we can identify the common points of the circle bundle of circles c

[1] Z

[2] Points {X

[1] Take a line (e) non intersecting the conic and consider its pole E.

[2] Find point Z

[3] Find points {X

[4] Determine points {E', E''} of the circle c

[5] Draw the circle c

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