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Art. LIV.—*On Orthogonal Circles*.

[*Read before the Philosophical Institute of Canterbury, 2nd December, 1914*.]

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The tripolar equation of the circle of radius *r*, which has its centre at the point O, whose trilinear co-ordinates are (α_{0}β_{0}γ_{0}), is
U≡aα_{0}X+bβ_{0}Y+cγ_{0}Z -2RS_{0} - 2#x0394S_{0} - 2#x0394r^{2} = 0 (i)

where S ≡ aβγ + bγα + cαβ and #x0394 is the area of the triangle of reference.^{*}

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Let *d* be the distance of O from H, the centre of the circle ABC. If U passes through H, then
R^{2}(αa_{0} + bβ_{0} + cγ_{0}) = 2RS_{0} +2 #x0394d^{2} *i.e*., 2RS_{0} = 2 #x0394 (R^{2} - d^{2});

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hence (i) may be written in the form U≡aα_{0}X+bβ_{0}Y+cγ_{0}Z =2#x0394 (R^{2} +r^{2} - d^{2}) (ii)

If *d*^{2} = R^{2} + *r*^{2}, the equation of the circle having its centre at (α_{0}β_{0}γ_{0}) and cutting the circle ABC orthogonally is
U≡aα_{0}X+bβ_{0}Y+cγ_{0}Z =0 (iii)

Its radius is given by the relation *r*^{2} = - R/#x0394 S_{o}, and its trilinear equation is
(aα +bβ +cγ){(cβ_{0}+bγ_{0})α +(aγ_{0}+cα_{0})β + (bα_{0} +aβ_{0})γ} = 2#x0394 (aβγ+bγα+cαβ).

From (iii) Ptolemy's theorem may be deduced. Let *r* = *o* so that U reduces to a point-circle lying on the circle ABC. Suppose its centre O to lie on the minor are BC of the circle ABC, and let D, E, F be the feet of the perpendiculars from O on BC, CA, AB respectively. We then have -2R.OD=OB.OC, 2R.OE=OC.OA, 2R.OF=OA.OB.

Substituting these for α_{0}β_{0}γ_{0} in the equation aα_{0}X_{0}+bβ_{0Y0}+cγ_{0}Z_{0} = *o*, we have AO.BO.CO[-BC.AO+CA.OB+AB.OC]=0, which is Ptolemy's theorem.

From the form of the equation U≡ aα_{0}X +bβ_{0}Y +cγ_{0}Z =0, it is seen that if U pass through the fixed point (X_{1}Y_{1}Z_{1}), the locus of the centre of the circles which pass through a fixed point and cut the circle ABC orthogonally is the straight line whose equation is X_{1}aα+Y_{1}bβ+Z_{1}cγ =c.

The equation of the orthogonal circle passing through two fixed points (X_{1}Y_{1}Z_{1}), (X_{2}Y_{2}Z_{2}) is |X Y Z X_{1} Y_{1} Z_{1} X_{2} Y_{2} Z_{2}|=0

[Footnote] * Trans. N.Z. Inst., vol. 46, p. 319.

Suppose the centre of the orthogonal circle to lie on the line pα+qβ+rγ= *o* at the point (λ/p,μ/q,ν/r), where λ + μ + ν = *o*. The equation of the circle will be λa/pX+μb/qY+νc/rZ=0, showing that the circle passes through the pair of inverse points determined by the intersections of the circles aX/p=bY/q=cZ/r.

Hence the theorem “All circles having their centres on a given line and cutting a given circle orthogonally pass through two fixed points which are inverse with respect to the given circle.”

Let A^{1}B^{1}C^{1} be a triangle inscribed in the circle ABC and circumscribed to the Brocard ellipse of the triangle ABC.

If λ + μ + ν = *o*, the trilinear ratios of A^{1}, B^{1}, C^{1} may be taken as (a/λ,b/μ,c/ν),(a/μ,b/ν,c/λ),(a/ν,b/λ,c/μ) respectively, for each of these ratios satisfies the equation of the circle ABC, and the line joining B^{1}C^{1} is α/aλ+β/bμ+γ/cν = *o*, the envelope of which is √a/α +√β/b +√γ/c = *o*.

Let three circles be described cutting the circle ABC orthogonally, each circle passing through two vertices of the triangle A^{1}B^{1}C^{1}. The centre of the circle through B^{1}C^{1} is the pole of B^{1}C^{1} with respect to the circle ABC, hence the trilinear ratios of the centre (α_{0}β_{0}γ_{0}) are [a(-1/λ+1/μ+1/ν), b(1/λ-1/μ+1/ν), c(1/λ+1/μ-1/ν)] and the tripolar equation of the circle is (-1/λ+1/μ+1/ν)a^{2}X+(1/λ-1/μ+1/ν)b^{2}Y+(1/λ+1/μ-1/ν)c^{2}Z=0 *i.e*., 1/λS_{1}+1/μS_{2}+1/νS_{3}=0,
where S_{1}≡-a^{2}X+b^{2}Y+c^{2}Z.

Hence the envelope of the system of orthogonal circles is √S_{1}+√S_{2}+√S_{3}=0

It may be remarked that S_{1} = *o* is the equation of the locus of points whose pedal triangles are right-angled, the vertex of the right angle lying on BC.

The radical centre of the three orthogonal circles is the symmedian point of the triangle ABC, for if L_{1} L_{3}≡α/aν+β/bλ+γ/cμ, then the radical axes of the circles in pairs are L_{2}-L_{3}=0, L_{3}-L_{1}=0, L_{1}-L_{2}=0, and it is at once seen that the equations of these lines are satisfied by the co-ordinates (*a, b, c*) of the symmedian point.

The locus of the centres of these circles is the conic whose equation is 1/β/b+γ/c+1/γ/c+α/a+1/α/a+β/b=0 This conic is the locus of points whose polars with respect to the triangle formed by the tangents to the circle ABC at the vertices A, B, C pass through the symmedian point of the triangle ABC.

To find the envelope of the circles cutting the circle ABC orthogonally and having their centres on the curve whose equation is
U≡(α^{2}/α^{2}-βγ/bc) ½+(β^{2}/b^{2}-γα/ca) ½)+(β^{2}/b^{2}-γa/ca) ½+(γ^{2}/c^{2}-αβ/ab) ½=0
Consider the points whose trilinear ratios are
(a/μ,b/v,c/λ),(a/v,b/λ,c/μ),where λ+μ=0.

These points lie on the circle ABC, and the equations of the tangents to the circle at them are
a/aμ^{2}=β/b^{2}+γ/cλ^{2}=0
a/aμ^{2}=β/b^{2}+γ/cλ^{2}=0
These lines will meet at the point
α:β:γ=α(λ;^{4})λ;^{4}-μ^{4})b(μ^{4}-v^{2}λ;^{2}:c(v^{7}-λ;^{2}μ^{2}),
which, since λ;^{2}-μv=μ^{2}-vλ;=v^{2}-λ;μ,reduces to α:β:γ=α(λ;^{2}+μ):b(μ^{2}+v^{2}λ;):c(v^{2}+λ;μ).
For the locus of the intersections of the tangents we have
γ/cλ;^{2}+a/aμ^{2}+β/bv^{2}=0 β/bλ;^{2}+γ/cμ^{2}a/av^{2}=0

giving
λ;^{2}:μ^{2}v^{2}=a^{2}/a^{2}-βγ/bc:β^{2}/b^{2}γa/ca:γ^{2}/c^{2}-αβ/ab,
and hence the locus of the intersection of the tangents is
U≡()a^{2}/a^{2}-βγ/bc) ½+β^{2}/b^{2}-γa/ca) ½+(γ^{2}/c^{2}-γ^{2}/c^{2}-αβ/ab) ½=0
The equation of the orthogonal circle may therefore be taken to be
α^{2}(λ;^{2}+μv)X=b^{2}(μ^{2}+vλ;)Y+c^{2}(v^{2}+λ;μ)Z=0;
or, writing λ;(μ+v),-μ(v+λ;)for λ;^{2},μ^{2}v^{2} respectively, the equation of the circle is
λ;v(-a^{2}X=b^{2}Y+c^{2}Z)+vλ;+λ;μ(a^{2}X+b^{2}Y-c^{2}Z)=0,
the envelope of which is
()-a^{2}X+b^{2}+c^{2}Z) ½(a^{2}X-b^{2}+c^{2}) ½+(a^{2}X-b^{2}Yc^{2}Z) ½+(a^{2}X+b^{2}Y-c^{2}Z) ½=0.

Let O be the pole of a conic S circumscribing the triangle ABC, and let P, Q be the extremities of a chord of the conic passing through O. If circles be drawn with their centres at P. Q and cutting the circle ABC orthogonally, they will intersect each other on the circle whose centre is at O and which cuts the circle ABC orthogonally.

Let S≡f/a+g/β+h/γ=0; and let α+μ+v=0 and λ;^{1=μ-v,μ2=vλ;,v1}=λ;-μ. We see by inspection that the two points whose trilinear ratios are (f/λ;,g,μ,h/v),(f/λ;^{1}g/μ^{1},h/v^{1}) lie on S. The equation of the chord PQ is
λ;λ;^{1}+μμ^{1}β/g+vv^{1}λ;/h=0,
which is satisfied by (*f, g, h*) – the co-ordinates of the pole of the conic S.

The circles having their centres at P, Q and cutting the circle ABC orthogonally will be S_{1}≡afX/λ;+bgY/μ+chZ/v=0 and for their intersections we have S_{2}≡afX/λ;_{1}+bgY/μ_{1}+chZ/v_{1}=0, since afX:bgY:chZ=1/μv^{1}-1/μ^{1}v:1/μ^{1}v:1/v^{1}λ;:1/v^{1}λ;:1/λ;μ^{1}1/λ;^{1}μ-1/λ;^{1}μ=λ;λ;^{1}:μμ^{1}:vv^{1}

sinceμv^{1}-μ^{1}v=vλ;^{1}v^{1}λ;=λ;μ^{1}-λ;^{1}μ=-σ(λ;^{2}).

Hence the locus of the intersection of S_{1} and S_{2} is afX+bgY+chZ=0 a circle having its centre at the pole (*f, g, h*) and cutting the circle ABC orthogonally.

If a circle cutting the circle ABC orthogonally meet the latter at the extremities of a chord passing through the symmedian point of the triangle ABC, it will pass through the two points whose pedal triangles are equilateral.

Any chord of the circle ABC which passes through the symmedian point (*a, b, c*) of the triangle ABC may be written
L≡λ;a/a+μβ/b-vγ/c=0
where λ;+μ+v=0

The co-ordinates of the pole of L with respect to the circle ABC are given by
α:β:γ=αλ;:bμcv;
hence the equation of the orthogonal circle is
λ;a^{2}X+μb^{2}Y+vc^{2}Z=0
Its form shows that it passes through the pair of inverse points given by
a^{2}X=b^{2}=c^{2}Z.

If P be one of the two points in the plane of the triangle ABC whose pedal triangles are equilateral, then AP sin A = BP sin B = CP sin C; hence the points determined by the above equations have their pedal triangles equilateral. It is at once seen that this pair of points lies on the Brocard diameter whose tripolar equation is
a^{2}(b^{2}-c^{2}X+b^{2}(c^{2}-a^{2}Y+c^{2}(a^{2}-b^{2}Z=0)

We now proceed to find the equation of the circle cutting the nine-point circle of the triangle ABC orthogonally.

Let D, E, F be the middle points of BC, CA, AB respectively, and let X_{1} = PD^{2}, Y_{1} = PE^{2}, Z_{1} = PF^{2}, where P is a point whose tripolar co-ordinates are (X, Y, Z), then
2X_{1}=Y+Z-a^{2}2 2Y_{1}=Z+X-b^{2}/2 2Z_{1}=X+Y-c^{2}/2

Let U ≡ - *a*_{α} + *b*_{β} + *c*_{γ} = *o*, V ≡ *a*_{α} - *b*_{β} + *c*_{γ} = *o*, W ≡ *a*_{α} + *b*_{β} - *c*_{γ} = *o*, be the equations of the sides of the triangle DEF, and let (α_{o}β_{o}γ_{o}) be the co-ordinates referred to the triangle ABC and (α_{1}β_{1}γ_{1}) the co-ordinates referred to the triangle DEF of the centre of a circle cutting the nine-point circle orthogonally, then
αα_{1}=U_{0},bβ=V_{0},cy_{1}=W_{0}.

The equation of the circle referred to the triangle DEF is
αα_{1}X_{1}+bβ_{1}Y_{1}+cγ_{1}Z_{1}=0
and this, when referred to the triangle ABC, becomes
U_{0}[Y+Z-a^{2}/2]+V_{0}[Z+X-b^{2}/2]+W_{0}[X+Y-c^{2}/2]=0; *i.e*., αα_{0}X+bβ_{0}Y+cy_{0}-¼(a^{2}U_{0}+b^{2}V_{0}+c^{2}W_{0}), which reduces to
αα_{0}X+bβ_{0}Y+cy_{0}=2R#x0394(α_{0}cos A+β_{0} cos B+γ_{0})

The envelope of circles having their centres on the circle ABC and cutting the nine-point circle of the triangle ABC orthogonally is α(X- ½bc cos A) ½b(Y- ½ ca cos B) ½+c(Z- ½ ab cos C) ½=0.

If circles be described cutting the nine-point circle orthogonally and having their centres at the vertices of a triangle inscribed in the circle ABC and circumscribed to the Brocard ellipse of the triangle ABC, their points of intersection will lie on the curve
1/a^{2}(X- ½ bc cos A)+1/b^{2}(Y- ½ cs cos B)+c^{2}(Z- ½ ab cos C)=0.

Let U_{1} be the result of substituting in (i) X_{1}Y_{1}Z_{1} for X, Y, Z respectively. Let a circle concentric with U = O pass through the point whose tripolar co-ordinates are (X_{1}Y_{1}Z_{1}), and let *r*' be the radius, of this circle, then
U_{1}≡α_{0}X_{1}+bβ_{0}c^{2}Y_{1}+cγ_{0}Z_{1}-RS_{0}-2#x0394r^{2}
and therefore U_{1} = 2 #x0394 (*r*'^{2} – *r*^{2}) = 2 #x0394 *t*^{2}, where *t* is the length of the tangent from (X_{1}Y_{1}Z_{1}) to the circle U = O.

Let *t*_{1}, *t*_{2}, *t*_{3} be the tangents to the circle *a*α_{o}X + *b*β_{o}Y + *c*γ_{o}Z = *o* from A, B, C respectively, then
2#x0394t^{1}_{2}=bβ^{2}c^{2}+cβ_{0}b^{2} 2#x0394t^{1}_{2}=aα_{0}c^{2}+cγa^{2} 2#x0394t^{1}_{2}aα_{0}t^{2}+bβ^{2}a^{2} 2#x0394=aα_{0}+bβ^{0}+cγ_{0}
Hence
t_{1}^{2}0c^{2}b^{2} t_{2}^{2}c^{2}0a^{2} t^{2}_{3}b^{2}a^{2}0 1111
which, on expansion, gives the following relation connecting the lengths of the tangents from the vertices of a triangle to any circle cutting orthogonally the circumcircle of the triangle, viz.,—
t_{1}^{2}sin2A+t_{2}^{2}sin 2B+t_{3}^{2}sin 2C=4#x0394.

If *t*′, *t*”′, *t*”′ be the lengths of the tangents to an orthogonal circle from the middle points of the sides of the triangle ABC, then
2#x0394t^{2}=aa_{0}m_{1}^{2}=aa_{0}m_{1}^{2}+a^{2}/4(bβ_{0}+cγ_{0})

2#x0394t′^{2}=bβ_{0}m_{2}^{2}+b^{2}/4(cγ+aa_{0})

2#x0394t′;^{2}=cγ_{0}m_{3}^{2}+c^{2}/4(aa_{0}+bβ).

where *m*_{1}, *m*_{2}, *m*_{3} are the medians of the triangle ABC. Hence
2#x0394(t′;^{2}+t′;^{2}+t′;^{2})=aa_{0}(m_{1}+b^{2}+c^{2}/4)+bβ(m_{2}^{2}+c^{2}+a^{2}/4)+cβ_{0}(m_{2}^{2}+a^{2}+b^{2}/4) =aa_{0}3(b^{2}+c^{2})-a^{2}+bβ_{0}3(c^{2}+a^{2})-b^{2}/4+cβ_{0}3(a^{2}+b^{2}-c^{2}/4)
also 2#x0394(t_{1}^{2}+t_{2}^{2}+t_{3}^{2})=aa_{0}(b^{2}+c^{2})+bβ_{0}(c^{2}+a^{2})+cγ_{0}(a_{0}+b^{2})
Hence
2#x0394{σ(t′^{2})-σ(t′;^{2})}=¼(aa_{0}+bβ_{0}+cγ_{0}σ(a^{2});
*i.e*.,
σ(t_{1}^{2})-(t′;^{2})=¼σ(a^{2}).