**Notation**

Let X* denote the isogonal conjugate of X.

Let DC(X) denote the circle with diameter XX*.

Let PC(X) denote the Pedal Circle of P.

Note that PC(X) = PC(X*), and that PC(X), DC(X) have the same centre.

Let T'(X) denote the *dual* of X, i.e. the tripolar of the isotomic conjugate of X.

Let C(X) denote the circumconic with perspector X.

Let F(X,Y) denote the fourth intersection of C(X) and C(Y).

**Preamble**

>From the Isocubics Book, an isogonal non-pivotal cubic can be described as either

(a) the locus of points X for which PC(X) is orthogonal to a fixed circle C(ω) whose centre is denoted by ω,

(b) the locus of points X for which DC(X) is orthogonal to a fixed circle C(Ω) whose centre is denoted by Ω.

We know that we have

*Fact 1*

ω is the mid-point of H and Ω.

Note that, if a fixed point P lies on the cubic, then the centre ω determines C(ω) as the unique circle with

centre ω orthogonal to PC(P).

*Construction A*

Let P' denote the mid-point of P and P*.

The circle on P'ω as diameter cuts PC(P) at points X,Y.

Then C(ω) is the circle centre ω through X (and Y).

Of course, similar remarks apply to Ω, C(Ω) and the circle DC(P).

Now suppose that we have points P and Q lie on the cubic, and that Q is not P or P*.

We have some obvious results :

*Fact 2*

The point ω lies on L(P,Q) the radical axis of PC(P) and PC(Q).

The point Ω lies on L'(P,Q) the radical axis of DC(P) and DC(Q).

It is easy to check that L(P,Q) and L'(P,Q) are parallel, and that H is twice as far from L'(P,Q) as from L(P,Q).

We are now in a position to investigate the problem using Cabri.

Given the points P and Q, then we can draw L(P,Q) and L'(P,Q).

Now *choose* a point ω on L(P,Q).

The corresponding Ω is determined by Facts 1 and 2.

Construction A now allows us to draw C(ω) and C(Ω).

Then we can draw the radical axis L(ω) of these circles.

*Cabri Observation*

The lines L(ω) all pass through a fixed point.

Note that the infinite point on L(P,Q) is also on L'(P,Q). Here, each "circle" is just the line joining the mid-points

of PP* and QQ*. These cannot coincide as they are the centres of inconics with diffferent real foci. Thus, if we

have the line L(ω) for a single finite ω, then the point identified by Cabri is the intersection of L(ω) with the line

of mid-points mentioned above.

We can then identify the circle C(P,Q) with this centre orthogonal to our C(ω), and hence also to C(Ω).

*Fact 3*

The circle C(P,Q) is the unique circle centred on the line of mid-points, and in the coaxal families generated by

(a) PC(P) and PC(Q), and by

(b) DC(P) and DC(Q).

**Theorem 1**

Suppose that we have points P and Q, with Q ≠ P, P*.

Each isogonal non-pivotal cubic may be described as a locus of type (a) or (b) above, where

(a) ω is on L(P,Q), and the circle C(ω) is orthogonal to C(P,Q),

(b) Ω is on L'(P,Q), and the circle C(Ω) is orthogonal to C(P,Q).

After much Maple calculation and Cabri sketching, we can identify C(P,Q). With hindsight, it is quite simple!

The starting point is a result about isogonal non-pivotal isocubics containing P and Q (and so also P* and Q*).

**Theorem 2**

Suppose that we have points P and Q, with Q ≠ P, P*.

The roots of isogonal non-pivotal isocubics containing P and Q lie on a line L"(P,Q).

The line L"(P,Q) contains the points F(P,Q), F(P,Q*), F(P*,Q), F(P*,Q*) (so these are collinear).

*Proof*

This can be verified by brute force, but we have an amusing alternative.

By *thinking* about the equation of nK(K,R,P), we see that the cubic contains Q if and only if R is on *some* line.

When the cubic degenerates it consists of the tripolar of R and its isogonal conjugate. If it contains P, P*, Q,

and Q*, and is degenerate, then the tripolar must contain P or P* **and** Q or Q*. In other words, R must be on

C(P) or C(P*) and on C(Q) or C(Q*), so that R is one of the quoted F(X,Y). It follows that these points lie on

the line of roots, and so are collinear!

We can easily write down an equation for the Pedal Circle of a point p:q:r. Suppose that this has the form

EPC(p,q,r,x,y,z) = 0. We also suppose that the circle on PP* as diameter has equation EDC(p,q,r,x,y,z) = 0.

In each case, we can choose equations with left-hand-side which is cubic in p,q,r and quadratic in x,y,z.

**Theorem 3**

Suppose that we have points P = p:q:r and Q = u:v:w, with Q ≠ P, P*.

Equations uvwEPC(p,q,r,x,y,z)-pqrEPC(u,v,w,x,y,z) = 0 and uvwEDC(p,q,r,x,y,z)-pqrEDC(u,v,w,x,y,z) = 0

define the *same* circle. This is our circle C(P,Q).

*Proof*

The fact that the two equations give the same circle C'(P,Q) is a matter of Maple computation.

>From the equations, C'(P,Q) belongs to the coaxal family **F**(P,Q) defined by PC(P) and PC(Q). Similarly,

it belongs to the coaxal family **G**(P,Q) defined by DC(P) and DC(Q).

Any circle (real or otherwise) orthogonal to PC(P) and PC(Q) is centred on L(P,Q) and passes through the

limit points of the family (which may not be real). It is therefore orthogonal to C'(P,Q).

Similarly, any circle orthogonal to DC(P) and DC(Q) is centered on L'(P,Q) and is orthogonal to C'(P,Q).

Thus C'(P,Q) contains the intersections of PC(P) and PC(Q), and those of DC(P) and DC(Q) since these

are "point-circles" orthogonal to the given pairs. Of course, these need not be real.

These intersections are also on our circle C(P,Q) - for the same reason - so C'(P,Q) = C(P,Q).

**Theorem 4**

Suppose that we have points P and Q, with Q ≠ P, P*.

In general, the centre of the circle C(P,Q) is the intersection of

(a) the line joining the mid-points of PP* and QQ*, and

(b) the line which is the dual of the infinite point of L"(P,Q).

**The case P = G**

The circle C(P,Q) has centre at the intersection of the lines joining

(a) X597 and the mid-point of P and P*

(b) G and the mid-point of the isotomic conjugates of the infinite points on T'(P) and T'(P*).

In the above, the second line may also be described as the line joining

(c) G and the barycentric product of the infinite points mentioned in (b).