Go to National Library of New Zealand Te Puna Mātauranga o Aotearoa
Volume 47, 1914
This text is also available in PDF
(285 KB) Opens in new window
– 584 –

Art. LIV.—On Orthogonal Circles.

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

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

The tripolar equation of the circle of radius r, which has its centre at the point O, whose trilinear co-ordinates are (α0β0γ0), is U≡aα0X+bβ0Y+cγ0Z -2RS0 - 2#x0394S0 - 2#x0394r2 = 0 (i)

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

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Let d be the distance of O from H, the centre of the circle ABC. If U passes through H, then R2(αa0 + bβ0 + cγ0) = 2RS0 +2 #x0394d2 i.e., 2RS0 = 2 #x0394 (R2 - d2);

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

hence (i) may be written in the form U≡aα0X+bβ0Y+cγ0Z =2#x0394 (R2 +r2 - d2) (ii)

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

If d2 = R2 + r2, the equation of the circle having its centre at (α0β0γ0) and cutting the circle ABC orthogonally is U≡aα0X+bβ0Y+cγ0Z =0 (iii)

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Its radius is given by the relation r2 = - R/#x0394 So, and its trilinear equation is (aα +bβ +cγ){(cβ0+bγ0)α +(aγ0+cα0)β + (bα0 +aβ0)γ} = 2#x0394 (aβγ+bγα+cαβ).

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Substituting these for α0β0γ0 in the equation aα0X0+bβ0Y0+cγ0Z0 = o, we have AO.BO.CO[-BC.AO+CA.OB+AB.OC]=0, which is Ptolemy's theorem.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

From the form of the equation U≡ aα0X +bβ0Y +cγ0Z =0, it is seen that if U pass through the fixed point (X1Y1Z1), 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 X1aα+Y1bβ+Z1cγ =c.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

The equation of the orthogonal circle passing through two fixed points (X1Y1Z1), (X2Y2Z2) is |X Y Z X1 Y1 Z1 X2 Y2 Z2|=0

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

– 585 –

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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 A1B1C1 be a triangle inscribed in the circle ABC and circumscribed to the Brocard ellipse of the triangle ABC.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

If λ + μ + ν = o, the trilinear ratios of A1, B1, C1 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 B1C1 is α/aλ+β/bμ+γ/cν = o, the envelope of which is √a/α +√β/b +√γ/c = o.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Let three circles be described cutting the circle ABC orthogonally, each circle passing through two vertices of the triangle A1B1C1. The centre of the circle through B1C1 is the pole of B1C1 with respect to the circle ABC, hence the trilinear ratios of the centre (α0β0γ0) 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)a2X+(1/λ-1/μ+1/ν)b2Y+(1/λ+1/μ-1/ν)c2Z=0 i.e., 1/λS1+1/μS2+1/νS3=0, where S1≡-a2X+b2Y+c2Z.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Hence the envelope of the system of orthogonal circles is √S1+√S2+√S3=0

It may be remarked that S1 = 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 section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

The radical centre of the three orthogonal circles is the symmedian point of the triangle ABC, for if L1 L3≡α/aν+β/bλ+γ/cμ, then the radical axes of the circles in pairs are L2-L3=0, L3-L1=0, L1-L2=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.

– 586 –

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

To find the envelope of the circles cutting the circle ABC orthogonally and having their centres on the curve whose equation is U≡(α22-βγ/bc) ½+(β2/b2-γα/ca) ½)+(β2/b2-γa/ca) ½+(γ2/c2-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.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

These points lie on the circle ABC, and the equations of the tangents to the circle at them are a/aμ2=β/b2+γ/cλ2=0 a/aμ2=β/b2+γ/cλ2=0 These lines will meet at the point α:β:γ=α(λ;4)λ;44)b(μ4-v2λ;2:c(v7-λ;2μ2), which, since λ;2-μv=μ2-vλ;=v2-λ;μ,reduces to α:β:γ=α(λ;2+μ):b(μ2+v2λ;):c(v2+λ;μ). For the locus of the intersections of the tangents we have γ/cλ;2+a/aμ2+β/bv2=0 β/bλ;2+γ/cμ2a/av2=0
giving λ;22v2=a2/a2-βγ/bc:β2/b2γa/ca:γ2/c2-αβ/ab, and hence the locus of the intersection of the tangents is U≡()a2/a2-βγ/bc) ½+β2/b2-γa/ca) ½+(γ2/c22/c2-αβ/ab) ½=0 The equation of the orthogonal circle may therefore be taken to be α2(λ;2+μv)X=b22+vλ;)Y+c2(v2+λ;μ)Z=0; or, writing λ;(μ+v),-μ(v+λ;)for λ;22v2 respectively, the equation of the circle is λ;v(-a2X=b2Y+c2Z)+vλ;+λ;μ(a2X+b2Y-c2Z)=0, the envelope of which is ()-a2X+b2+c2Z) ½(a2X-b2+c2) ½+(a2X-b2Yc2Z) ½+(a2X+b2Y-c2Z) ½=0.

– 587 –

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.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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;1g/μ1,h/v1) lie on S. The equation of the chord PQ is λ;λ;1+μμ1β/g+vv1λ;/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 S1≡afX/λ;+bgY/μ+chZ/v=0 and for their intersections we have S2≡afX/λ;1+bgY/μ1+chZ/v1=0, since afX:bgY:chZ=1/μv1-1/μ1v:1/μ1v:1/v1λ;:1/v1λ;:1/λ;μ11/λ;1μ-1/λ;1μ=λ;λ;1:μμ1:vv1

sinceμv11v=vλ;1v1λ;=λ;μ1-λ;1μ=-σ(λ;2).

Hence the locus of the intersection of S1 and S2 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.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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 section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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;a2X+μb2Y+vc2Z=0 Its form shows that it passes through the pair of inverse points given by a2X=b2=c2Z.

– 588 –

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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 a2(b2-c2X+b2(c2-a2Y+c2(a2-b2Z=0)

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

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Let D, E, F be the middle points of BC, CA, AB respectively, and let X1 = PD2, Y1 = PE2, Z1 = PF2, where P is a point whose tripolar co-ordinates are (X, Y, Z), then 2X1=Y+Z-a22 2Y1=Z+X-b2/2 2Z1=X+Y-c2/2

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Let U ≡ - aα + bβ + cγ = o, V ≡ aα - bβ + cγ = o, W ≡ aα + bβ - cγ = o, be the equations of the sides of the triangle DEF, and let (αoβoγo) be the co-ordinates referred to the triangle ABC and (α1β1γ1) the co-ordinates referred to the triangle DEF of the centre of a circle cutting the nine-point circle orthogonally, then αα1=U0,bβ=V0,cy1=W0.
The equation of the circle referred to the triangle DEF is αα1X1+bβ1Y1+cγ1Z1=0 and this, when referred to the triangle ABC, becomes U0[Y+Z-a2/2]+V0[Z+X-b2/2]+W0[X+Y-c2/2]=0; i.e., αα0X+bβ0Y+cy0-¼(a2U0+b2V0+c2W0), which reduces to αα0X+bβ0Y+cy0=2R#x0394(α0cos A+β0 cos B+γ0)

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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/a2(X- ½ bc cos A)+1/b2(Y- ½ cs cos B)+c2(Z- ½ ab cos C)=0.

– 589 –

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Let U1 be the result of substituting in (i) X1Y1Z1 for X, Y, Z respectively. Let a circle concentric with U = O pass through the point whose tripolar co-ordinates are (X1Y1Z1), and let r' be the radius, of this circle, then U1≡α0X1+bβ0c2Y1+cγ0Z1-RS0-2#x0394r2 and therefore U1 = 2 #x0394 (r'2r2) = 2 #x0394 t2, where t is the length of the tangent from (X1Y1Z1) to the circle U = O.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

Let t1, t2, t3 be the tangents to the circle aαoX + bβoY + cγoZ = o from A, B, C respectively, then 2#x0394t12=bβ2c2+cβ0b2 2#x0394t12=aα0c2+cγa2 2#x0394t120t2+bβ2a2 2#x0394=aα0+bβ0+cγ0 Hence t120c2b2 t22c20a2 t23b2a20 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.,— t12sin2A+t22sin 2B+t32sin 2C=4#x0394.

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

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#x0394t2=aa0m12=aa0m12+a2/4(bβ0+cγ0)

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

2#x0394t′2=bβ0m22+b2/4(cγ+aa0)

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

2#x0394t′;2=cγ0m32+c2/4(aa0+bβ).

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

where m1, m2, m3 are the medians of the triangle ABC. Hence 2#x0394(t′;2+t′;2+t′;2)=aa0(m1+b2+c2/4)+bβ(m22+c2+a2/4)+cβ0(m22+a2+b2/4) =aa03(b2+c2)-a2+bβ03(c2+a2)-b2/4+cβ03(a2+b2-c2/4) also 2#x0394(t12+t22+t32)=aa0(b2+c2)+bβ0(c2+a2)+cγ0(a0+b2) Hence 2#x0394{σ(t′2)-σ(t′;2)}=¼(aa0+bβ0+cγ0σ(a2); i.e., σ(t12)-(t′;2)=¼σ(a2).