## center of general circle

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• If a circle is written in this form w S0 -(Px+Qy+Rz)(x+y+z) = 0 where S0 is the equation of the circumcircle, its center can be written ... Since I had never
Message 1 of 11 , Apr 5, 2000
If a circle is written in this form

w S0 -(Px+Qy+Rz)(x+y+z) = 0

where S0 is the equation of the circumcircle, its center can be written

: bb Q - SA R - SC P - bb SB w :

Since I had never seen this and it is very simple, I thought others
might be interested.

Steve
• ... This is one of several formulae that I can never remember, and have had to painfully reconstruct each time I use it. In fact I ve been intending to do
Message 2 of 11 , Apr 6, 2000
On Wed, 5 Apr 2000, Steve Sigur wrote:

> If a circle is written in this form
>
> w S0 -(Px+Qy+Rz)(x+y+z) = 0
>
> where S0 is the equation of the circumcircle, its center can be written
>
> : bb Q - SA R - SC P - bb SB w :
>
> Since I had never seen this and it is very simple, I thought others
> might be interested.

This is one of several formulae that I can never remember, and have
had to painfully reconstruct each time I use it. In fact I've been
intending to do just that "in the next few days" for several weeks now!

Let me rephrase, using my standard notation, and make a few remarks.
I use the notation (P:Q:R|w) for the equation

(pX + qY + rZ).sigma = phi,

where sigma(X,Y,Z) = X + Y + Z and phi(X,Y,Z) = aaYZ + bbZX + ccXY.
As Steve kindly points out, the center of this circle is

( : pSC - qbb + rSA + wbbSB : ).

Perhaps the most important of the "other formulae" that I'm thinking of
are that for the radius of this circle, and the condition for two such
circles to be orthogonal.

Let me help (myself at least) to remember this one. On the one hand,
if p = q = r = 0 we must get the circumcenter (:bbSB:) - this will
remind us of the coefficient of w. On the other hand, I am very
familiar indeed with the fact that the vector (SC:-bb:SA) is
perpendicular to the edge b. I don't quite see just why these should
be the coefficients of p,q,r; but at least this fact can be remembered.

Now let me help everyone else to find equations of circles. The key
is that if one normalizes to the form (p:q:r|1) which has w = 1, then
the other coefficients p, q, r are the powers of A, B, C with respect
to the circle. Recall that the power of a point P with respect to
some circle is the (properly signed) product PX.PY for any line PXY
that cuts the circle in X and Y. In particular, it equals PT^2
if PT is tangent to the circle at T.

This makes it astonishingly easy to find and remember the equations of
all sorts of interesting circles. For instance, A,B,C are on the
circumcircle, while the tangents from them to the incircle have lengths
sa,sb,sc, so these circles have the equations

(0:0:0|1) and (sa^2:sb^2:sc^2|1).

Again, for the adjoint circle that passes through A and C and
touches the edge a, the powers of A,B,C are obviously 0,aa,0,
and so this circle has equation (0:aa:0|1).

While I'm at it, I think I'll just list some useful circles.
We've often discussed what I call the "Eulerian" coaxal system,
generated by the circumcircle and Ninecircle. The "Eulerian" circle
centered at the typical point Ep = O + p(H-O) of the Euler line
is (pSA:pSB:pSC|1) ~ (SA:SB:SC|1/p). In particular, this gives:

(0:0:0|1) (SA:SB:SC|1) (SA:SB:SC|2) & (SA:SB:SC|3/2)
for the
circum- polar Nine- & orthocentroidal
circles.

Another well-known family of circles are the Cevian circles,
ie, those based on Cevians as diameters. The typical B-Cevian
circle has the form (pSA:0:sSC|p+r) (so these circles form
a coaxal system), and in particular we have the edge-circles
(SA:0:0|1) &c and the altitude circles (SA:0:SC|1) &c. All
these circles belong to the linear system (pSA:qSB:rSC|p+q+r)
that includes the Ninecircle.

Steve has recently mentioned the Tucker circles. The typical
one pases through the points marked in the figure, wherein the
long edges of the inscribed (crossed) hexagon are parallel to
the sides and the short ones antiparallel:

B
/ \
(0:1-cct:cct) *---* (aat:1-aat:0)
/ \ / \
/ X \
(0:bbt:1-bbt) *---/-\---* (1-bbt:bbt: 0)
/ \ / \ / \
C---*-----*---A
/ \
(aat:0:1-aat) (1-cct:0:cct)

Its equation, if I remember it correctly, is

( bbcct - aabbcctt : ccaat - aabbcctt : aabbt - aabbcctt | 1 ).

= ( p : q : r | 1), say.

Let's use the formula Steve gives to find its center:

( : wbbSB + pSC - qbb + rSA : ) =

( : bbSB + t[bbccSC-bbccaa+aabbSA] - ttaabbcc(SC-bb+SA) : )

= ( : bbSB + tbb[aabb+bbcc-aaaa-cccc] : )

= ( : bbSB + tbbbb[aa+bb+cc] - tbb[aaaa+bbbb+cccc] : )

which expresses it as a linear combination of

(:bbSB:) = O, (:bbbb:) = pK, (:bb:) = K,

and so shows that it moves along the Brocard meridian line m.

I started a calculation to locate it more exactly along m,
but it got a bit to complicated, so I'll do it on paper and
report the result later.

Regards to all, JHC
• ... This prompts a few comments. The first is that since x,y,z are commonly used for (orthogonal) trilinear coordinates, I think it best to call the
Message 3 of 11 , Apr 6, 2000
On Wed, 5 Apr 2000, Steve Sigur wrote:

> John's spiffy circle formula is really
>
> S0 - (Px +Qy + Rz)(x+y+z) = 0

This prompts a few comments. The first is that since x,y,z are
commonly used for (orthogonal) trilinear coordinates, I think it best
to call the barycentric coordinates X,Y,Z (this convention also allows
us to use both systems when that's convenient). Correspondingly, I'll
switch to small letters p,q,r,s for the coefficients, to keep them
looking different to the coordinates.

The second is that there are quite a few advantages to writing the
equation in the form

pX + qY + rZ - s.phi/sigma = 0,

where as always phi = aaYZ+bbZX+ccXY and sigma = X+Y+Z,
since when s = 0 this is just the standard equation of a straight line.

So I've been thinking of introducing a suitable notation, say

(p:q:r|s)*(X:Y:Z)

for the above expression, which is a sort of non-linear inner product.
When s = 0, I usually omit it, giving the notation

(p:q:r|) for pX + qY + rZ = 0,

the equation of the general straight line.

The third comment is that in view of their appearance in such
equations, when tabulating coordinates of various points, it's a
good idea always to tabulate phi and sigma, and maybe phi/sigma,
which is sometimes simpler than phi or sigma. A good way to do
this is to write the fraction with numerator phi and denominator
sigma, and then simplify when possible. For example

G = (1:1:1) (aa+bb+cc):3

H = (:SCA:) SS.SABC:SS = SABC

O = (:bbSB:) aabbccSS:2SS = aabbcc/2

K = (:bb:) 3aabbcc:aa+bb+cc

bB = (cc:bb:aa) bb(aaaa+ccaa+cccc):aa+bb+cc

bB' = (aa:2SB:cc) ccaa(2aa+2cc-bb):(2aa+2cc-bb) = ccaa

in which I've used the form N:D to indicate that fractions are being
used in this special way.

> Lemoine's parallel circle
> 4SS/aabbcc S0 - ((bb+cc)/aa x + (cc+aa)/bb y + (aa+bb)/cc z)(x+y+z)=0
>
> SS/aabbcc S0 - (SA/aa x + SB/bb y + SC/cc z)(x+y+z) = 0

It's well worth while learning and using the (p:q:r|s) notation.
What you're saying is that Lemoine's parallel circle is

( : (cc+aa)/bb : | SS/aabbcc ) ~ ( : ccaa(cc+aa) : | SS )

and I presume the second one is his antiparallel circle:

( : SB/bb : | SS/aabbcc ) ~ ( : ccaaSB : | SS ).

I suspect you may be worried, Steve, about burdening people's
memories with special notations (as I am, since I've introduced
so many of them!). But if someone is going to be working at all
with equations of circles, they'll HAVE to learn the general equation

(pX+qY+rZ)(X+Y+Z) = s(aaYZ+bbZX+ccXY),

and abbreviating this to (p:q:r|s) will be a relief rather than
a strain.

JHC
• Recent postings by Paul Yiu and John Conway have discussed Lucas circles. These are circles Cx where Ca touches Cb, Cc and the circumcircle (at A). Paul gave
Message 4 of 11 , Apr 9, 2000
Recent postings by Paul Yiu and John Conway have discussed Lucas circles.
These are circles Cx where Ca touches Cb, Cc and the circumcircle (at A).
Paul gave the perspectors of the desmic configuration determined by the
mutual points of contact. I write these perspectors as L1/2, L1, L0
where Lk=[aa(Sa+kS)::], the desmon L1/2 being the radical centre, L0 the
circumcentre and (as John pointed out) L1 is the isogone of the Vecten
point Vn. I note that the desmic mates in this case are
Qa=[aa(Sa+S):bbSb:ccSc] etc, so that Qa divides OA in the ratio S+aa:S.
This is the ratio of the radii of of the circumcircle and Lucas circle Ca,
so that the Qx are the homothetic centres of the circumcircle with each
Lucas circle

Another desmic configuration is that determined by the centres of the Lucas
circles, Oa=[aa(Sa+2S):bbSb:ccSc] etc. This has perspectors L1, L2, L0 -
the desmic mates being the intersections BOc.COb etc.

These two desmic configurations may be written as D1 and D2 where Dk has
perspectors Lk/2, Lk, L0. This indicates a generalisation of Lucas circles
that replaces S by kS to yield new circles Cx which touch the circumcircle
at A,B,C, with the other homothetic centres (with the circumcircle) being
the previous mutual points of contact of the Lucas circles. The desmon
Lk/2 is (as before) the radical centre. And so on. It may be worth noting
that when k<1, the generalised circles intersect in pairs and that these
intersections yield some interesting triangles.

More generally still, the points Li, Lj, Lk will be the perspectors of a
desmic configuration (with the first as desmon) when i=(j+k)/2. Note
that L(+-1/root3) are the isodynamic points and that L(+-root3) are
isogones of the napoleon points....

Finally, I wonder whether there could be some single noun to denote phrases
like a desmic confuguration, a desmic system, a desmic set of three
tetrads, etc. The dictionary gives a noun DESMA, plural DESMATA. Would
DESM be acceptable?

Dick Tahta
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