Here we discuss the usual centre, i.e the pole in C(P,Q) of the Line at Infinity.

As usual, we write P = p:q:r, Q = u:v:w.

The coordinates of the centre are easily obtained from its matrix. They are complicated.

In **[BG]**, Bernard looks at various associated problems :

(1) There is a unique Q such that C(P,Q) has centre P,

(2) The centre of C(P,Q) is on PQ if and only if G is on the circumcubic through P and Q.

(3) There are three points with the mid-point of PQ the centre of (non-degenerate) C(P,Q).

There is one other case with a nice answer :

**Theorem 1**

The centre of C(P,Q) lies on T(P) if and only if Q lies on the line Σ qr(2p+q+r)x = 0.

Here we look at the problem of determining when the centre of C(P,Q) is on a fixed line.

In general, the answer is that Q is on a quadric depending on P and on the line. In fact,

the isotomic conjugate is on a conic, but we do not find it useful, except when the line is

the line at infinity. Then C(P,Q) is a parabola. This is discussed in **[BG]**.

To make progress, we further restrict the problem to that of finding C(P,Q) within a given

homothety class, and having its centre on a given line. The homothety class is considered

in **[WWS]**. There we saw that C(P,Q) is homothetic with the circumconic C(S), where

S =[ pu(q+r)(v+w)]. If S is fixed, then we can express the coordinates of Q in terms of

those of P. If we also fix the line on which the centre lies, we get a condition on P.

**Note**

If the centre X is the infinite point of the given line, then C(P,Q) is a parabola. As this conic

is homothetic to C(S), this must also be a parabola with the same centre. Then S must be

the barycentric square of X, and so is on the Steiner Inellipse. Below, we exclude the cases

where S is on the Steiner Inellipse, or the line is the line at infinity.

The calculation puts P on a sextic, but this is the isotomic conjugate of a cubic. In turn,

the cubic is the anticomplement of a circumcubic which is of type nK, with pole S, and

root on the Line at Infinity.

To make the expression neater, we introduce the maps

ta(M) is the isotomic conjugate of the anticomplement of M,

ct(M) is the complement of the isotomic conjugate of M.

Note that ta(M) is the perspector of the inconic with centre M.

Then C(P,Q) is homothetic with C(S) if and only if Q is ta(S-isoconjugate of ct(P)).

**Note**

If S = X^{2}, with X at infinity, so S is on the Steiner Inellipse, then for any P, and Q as

above. Then C(P,Q) will be a parabola with centre X as it is homothetic with C(S).

We shall write nK(S,R,t) for the cubic of type nK with pole S, root R, and with t the

coefficient of xyz in the equation where the other coefficients are derived from the

barycentric coordinates of S and R. We may also write nK(S,R,M) for the cubic of

type nK with pole S root R, passing through M.

**Theorem 2**

Let S be the point k:l:m, not on the Stener Inellipse, and

**L** the line Kx+Ly+Mz = 0, not the line at infinity.

Let L" be the infinite point of **L**.

(1) C(P,Q) is homothetic with C(S) and has centre on **L** if and only if

ct(P) and ct(Q) are S-isoconjugates on nK(S,R,t) (R,t given in the appendix).

R is the infinite point on the polar of L" in C(S). See the remark below.

(2) the cubic is of type nK0 if and only if the centre of C(S) is on **L**, and then

R is the L'-Ceva conjugate of S, L' the tripole of **L**.

**Corollary**

Let S be a point not on the Steiner Inellipse, and X a finite point not fixed by S-isoconjugation.

Then there are three pairs {P,Q} such that C(P,Q) is homothetic with C(S), and has centre X.

*Proof Notes*

If we choose two lines through X, then ct(P) must lie on the respective nK(S,R,t).

Apart from A,B,C, there are three isoconjugate pairs on the intersection of the cubics.

Observe that the formulae in the Appendix show that the cubics arising from the lines

through X form a pencil through the above points.

The restriction on S is, as usual, needed to ensure that C(S) is not a parabola.

The restriction on X emerges from the following Lemma.

**Lemma**

The inconic with centre X is homothetic with C(S) if and only if X is fixed by S-isoconjugation.

*Proof Notes*

An inconic I is homothetic with the circumconic C(X^{2}), where X is the centre of I.

When X is fixed by S-isoconjugation, the inconic with centre X arises as C(P,P), with P = ta(X).

Then the corresponding cubics are of type cK(S,R), with R at infinity. The point X accounts for

four of the intersections, so there is just one other pair of intersections. Calculation shows that

the roots are real precisely when X is outside the Steiner Inellipse.

We now consider the cases excluded from Theorem 2 and its Corollary.

If **L** is the line at infinity, so we may take K = L = M =1, then the equations

in the appendix give "R" = 0:0:0, so the "cubic" consists of the sidelines or

is the entire plane. In the former case, the C(P,Q) are parallel line pairs.

In the other case, we get the family of parabolic C(P,Q) as before.

If S is on the Steiner Inellipse, then the "cubic" degenerates into the

sidelines of the anticevian triangle of X, where S = X^{2}, X at infinity.

Then the corresponding C(P,Q) degenerate into line pairs meeting at X.

**Remark**

The root R was identified as the infinite point on the polar of L" in C(S).

This is a generalization of the orthopoint. When S = X(6), we get the usual orthopoint.

R is also the infinite point of the line joining ta(S) and the K-isoconjugate of a(S), the

anticomplement of S.

Obviously, if we choose M on the cubic nK(S,R,t), with isoconjugate M*, then C(ta(M),ta(M*))

is homothetic with C(S). Theorem 2 shows that we will have the centre on a fixed line if we

take R on the line at infinity.

**Theorem 3**

Suppose that S is a fixed point, not on the Steiner Inellipse, and

R is a fixed point at infinity.

Let **K**(t) = nK(S,R,t).

If M is on **K**(t), then C(ta(M),ta(M*)) is homothetic with C(S).

(1) If t = 0, the centre is on **L**(0), the polar of R in C(S).

Or, equivalently, the tripolar of the cevapoint of S and R.

(2) In general, the centre is on a line **L**(t), parallel to **L**(0).

Each line in (2) is the polar in C(S) of a point on the line joining R to the centre of C(S).

The algebra requires us to make the restriction on S.

In general, we get equations for suitable K,L,M from the coordinates of R and

the constant t as in the Appendix. As R is at infinity, these will have a unique

solution provided that S is not on the Steiner Inellipse.

Suppose that S is on the Steiner Inellipse, so S = U^{2}:V^{2}:W^{2}, with U+V+W = 0.

If we try to identify the K,L,M giving rise to nK(S,R,t), then we get inconsistent

equations unless R = xU:xV:xW, the centre of C(S), and t = 2xUVW. Then the

cubic nK(S,R,t) is degenerate. The corresponding C(P,Q) also degenerate, each

into a pair of lines through R.

In Theorem 1, we discussed the case where that centre of C(P,Q) in on PQ. If we add the

condition that C(P,Q) is homothetic with a given C(S), then we get a nice result.

**Theorem 4**

Suppose that S is a fixed point, not on the Steiner Inellipse.

Then C(P,Q) is homothetic with C(S) and has its centre on PQ if and only if

P (and hence Q) lie on pK(X2,aS), where aS is the anticomplement of S.

For such P, Q = ta(ct(P)*), where X* denotes S-isoconjugation.

For such P,Q, PQ passes through ataS, which also lies on pK(X2,aS).

Notes

The homothety condition here excludes the cases where C(P,Q) is degenerate. It then follows

that the condition that the centre is on PQ is that G is on the circumconic through P,Q.

The description of Q in terms of P comes from Theorem 2, so Q is on the cubic.

ataS is the aS-Ceva conjugate of X2, so must lie on the cubic as X2 does.

When S = X6, the cubic is K007, the Lucas Cubic, and ataS = X20.

**The case S = X(6)**

These give the circular C(P,Q). The nK are isogonal. The nK0 occur when X(3) is on **L**.

K018 = nK0(X(6),X(523)).

Circular C(P,Q) with centres on the Euler Line are associated with nK0(X(6),X(523)),

since X(523) is the orthopoint of X(30), the infinite point on the Euler Line.

The Nine-point Circle derives from {X(2),X(6)}.

We also get circles from {X(13),X(15)}, and {X(14),X(16)} among others.

K325 = nK0(X(6),X(512)).

Circular C(P,Q) with centres on the Brocard Axis are associated with nK0(X(6),X(512)),

since X(512) is the orthopoint of X(511), the infinite point on the Brocard Axis.

There are no known triangle centres on K325, but it does contain intersections of

medians and symmedians. Cabri confims that these give rise to circular C(P,Q) with

centres on the Brocard Axis. There is a connection with Bernard Gibert's **[BG2]**.

K067 = nK(X(6),X(30),X(2)).

This contains {X(2),X(6)} so the line of centres contains X(5).

It also contains X(523), the orthopoint of X(30), so is the line X(5)X(523).

The cubics nK(X(6),X(30),t) give rise to circular C(P,Q) with centres on lines containing

X(523). These include the Orthic and de Longchamps Axes.

K086 = cK(#X(1),X(514))

This cubic contains X(1), and ta(X(1)) = X(7), so the family of C(P,Q) contains the incircle.

The line **L** also contains X(516), the orthopoint of X(514), so is the line X(1)X(7) as this

contains X(516).

K137 = cK0(#X(1),X(513)).

As an nK0, the associated line **L** is X(1)X(3) as the line through X(3) and X(517), the

orthopoint of X(513). Or since the line contains X(1) as above, as well as X(3), the centre

of C(X(6)).

In general, the line **L** for cK(#X(1),R), with R at infinity is the line joining X(1) to the

orthopoint of R. The only cK0 in the class is K137 since the line must contain X(3).

K187 = nK(X(6),X(525),X(3)).

Maple calculation shows that the centres of associated C(P,Q) lie on X(20)X(64).

The cubic contains the isogonal pairs {X(3),X(4)}. Applying the map ta, we get

{P,Q} = {X(69),X(253)}. Note that these lie on the line of centres. After 2.3.1(1),

we know that we must have X(2) on the circumconic through X(69) and X(253).

The coordinates of the centre are rather complicated.

The cubic also contains the foci of the Steiner Inellipse. Then we have a pair {P,Q}

consisting of the isotomic conjugates of the foci of the Steiner Ellipse. Thus, we see

the isotomic conjugates of these latter foci are cyclocevian conjugates.
Here, Cabri

shows that the centre is X(20).

K248 = nK(X(6),X(512),X(187)).

Maple calculation shows that the centres of the associated C(P,Q) lie on X(20)X(185).

This can also be deduced as follows. As the foci of the Steiner Inellipse lie on the cubic,

the results for K187 show that the line contains X(20). The general theorey shows that

it passes through X(511), the orthopoint of X(512), so the line is parallel to the Brocard

Axis. The cubic also contains the isgonal conjugate pairs {Brocard points), and also

(X(187),X(671)}. These give rise to circular C(P,Q), The centres appear complicated.

K352 = nK(X(6),X(522),X(9)).

Maple calculation shows that the centres of the associated C(P,Q) lie on X(20)X(8).

This also follows as this is the line through X(20) and X(515), the orthopoint of X(522).

The cubic contains the foci of the Steiner and Mandard Inellipses, but the related

centres are complicated. The pair {X(9),X(57)} leads to the circular CP(X(8),X(189)),

with centre X(1158) on X(8)X(20). Of course, X(1158) is already known to be the

circumcentre of the extouch triangle - the cevian triangle of X(8).

**The case S = X(2)**

These give C(P,Q) homothetic with the Steiner Ellipse.

The nK0(X(2),R) with R at infinity are actually cK(#X(2),R), so constitute CL031.

In these cases, We have X(2) on **L**, and then R = K:L:M.

The only examples in CTC are K185 and K296.

In general, we get nK(X(2),R,t), with R = -2K+L+M:K-2L+M:K+L-2M, t = 2(K+L+M).

Note that R is the infinite point on the line joining X(2) and K:L:M.

**A final example**

K068 = nK0++(X(523),X(524)).

The conic C(X(523)) is the Kiepert Hyperbola **KH**. Our theory gives C(P,Q) which are

homothetic with **KH**. The centres lie on the polar of X(524) in **KH**, the line X(2)X(99).

The known centres{X(2),X(523)} on K068 give the rectangular hyperbola C(X(2),X(99)).

This is the complement of the Kiepert Hyperbola - as P = X(2).

The centre is X(620) - the complement of X(115), the centre of the Kiepert Hyperbola.

It contains X(2) - as X(99) is on C(X(2)), the Steiner Ellipse. See Corollary 2 of **[BG]**.

It contains X(3) - as the complement of X(4), or as X(4) of the cevian triangles of X(2),

and X(99). The latter is a property of points on the Circumcircle. See also 3.2.3 of **[BG]**.

It contains X(2482) - the complement of X(671), the intersection of the Kiepert Hyperbola

and the Steiner Ellipse. This is on X(2)X(99), so the centre X(620) is the mid-point of X(2)

and X(2482). Of course, X(2482) is on the Steiner Inellipse.

It contains X(114) as the complement of X(98), the intersection of the Kiepert Hyperbola

and the Circumcircle. This is also the reflection of X(3) in the centre.

It contains X(1649) as the X(99)-Ceva conjugate of X(524) - T(X(2))nT(X(99)). This is the

fourth intersection with I(X(99)).

**Appendix**

The coordinates of the root R in Theorem 2 are given by

x = -2Kk+L(k+l-m)+M(m+k-l),

y = K(k+l-m)-2Ll+M(l+m-k),

z = K(m+k-l)+L(l+m-k)-2Mm.

Note that x+y+z = 0.

The constant t in Theorem 2 is given by

t = 2(Kk(-k+l+m)+Ll(k-l+m)+Mm(k+l-m)).

**[WWS]** nine-point conic notes

**[BG]** Bernard Gibert's Bicevian Conics

**[BG2]** Tucker Circles.