If in the above equation [R cos A, R cos B, R cos C] be substituted for (a₀β₀γ₀) and ρ be made equal to R, we obtain for the equation of the circle ABC

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

a cos A X + b cos B Y + c cos B Z − abc = o (3)

If the square of (3) be subtracted from the fundamental relation connecting four coplanar points, viz.,

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

σ(a²X²−2σ(bc cos A YZ) − 2 abc σ (a cos A X) + a²b²c² = 0, (4)

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

we have σ(a⁴X²)−2σ(b²c²YZ) = 0,

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

whence (aX^½+bY^½+cZ^½) (−aX^½+bY^½+cZ^½) (aX^½−bY^½+cZ^½) (aX^½+bY^½−cZ^½)=0,

a known relation connecting the distances of any point on a circle from the vertices of an inscribed triangle.

The following list contains the equations of some of the more important circles connected with the triangle.

1. Centre at circumcentre, radius = ρ

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

a cos A X + b cos B Y + c cos C Z = 4R sin A sin B sin C (R² + ρ²).

2. Centre at orthocentre, radius = ρ

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

tan A X + tan B Y + tan C Z = ρ² tan A tan B tan C + 4δ

In-circle of pedal triangle, ρ = 2R cos A cos B cos C

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

tan A X + tan B Y + tan C Z = ½R² sin 2A sin 2B sin 2C + 4δ.

Polar circle, ρ² = − 4R² cos A cos B cos C

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

tan A X + tan B Y + tan C Z = 2δ.

3. Centre at centroid, radius = ρ

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

X + Y + Z = ⅓σ (a²) + 3σ².

4. Centre at symmedian point, radius = ρ

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

a²X+b²Y+c²Z=3a²b²c²/σ(a²) + σ(a²ρ².

5. Centre at in-centre, radius = ρ

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

aX+bY+cZ=(a+b+c) (2Rr+ρ²).

In-circle, ρ = r

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

aX − bY − cZ = (b + c − c) (2Rr + r²).

6. Centre at ex-centre (−111), radius = ρ

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

aX − bY − cZ = (b + c − a) (2Rr₁ − ρ²).

Ex-circle, ρ = r₁

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

aX − bY − cZ = (b + c − a) (2Rr₁ − r₁²).

7. Circle concentric with circle I₁I₂I₃, radius = ρ

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

a (2R − r₁) X + b (2R − r₂) Y + c (2R − r₃) Z = abcr + 2°ρ². Circle I₁I₂I₃, ρ = 2R

a (2R − r₁) X + b (2R − r₂) Y + c (2R − r₃) Z = abc (2R + r).

8. Centre at Nine-point centre, radius = ρ

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

R [a cos (B − C) X + b cos (C − A) Y + c cos (A − B) Z] = R²ρ (3 + 8 cos A cos B cos C) + 4δρ².

Nine-point circle, ρ = ½ R.

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

a cos (B − C) X + b cos (C − A) Y + c cos (A − B) Z = abc (1 + 2 cos A cos B cos C).

Orthocentric circle, ρ² = R²/4 (1 − 8 cos A cos B cos C)

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

a cos (B − C) X + b cos (C − A) Y + c cos (A − B) Z = abc.

9. Circles of Appolonius

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

b²Y − c²Z = o, c²Z − a²X = o, a²X − b²Y = o.

10. Circles with centres at Brocard points

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

Ω(c/b, a/c, b/a), Ω'(b/c,c/a,a/b), radius=ρ c²a²X+a²b²Y+b²c²Z = a²b²c² + σ (b²c²)ρ² b²a²X+c²b²Y+a²c²Z = a²b²c² + σ (b²c²)ρ²

11. Circle having ΩΩ' as diameter

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

a² (b² + c²) X + b² (c² + a²) Y + c² (a² + b²) Z = a²b²c² (1 + 2 cos 2°),

where ω is the Brocard angle of the triangle.

12. Brocard circle

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

abc σ (X) + σ (a³ cos A X) = ½abc σ (a²).

13. Circle on II₁ as diameter

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

− a²X + (b + c) (bY + cZ) = a²bc.

14. Circle on I₂I₃ as diameter

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

a²X − (b − c) (bY − cZ) = a²bc.

15. Circle having the bisector of the angle A as diameter

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

(b + c) X + bY + cZ = bc (b + c).

16. Circle having side of pedal triangle as diameter

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

a²sec A cos (B − C) X + b² + c²Z = 8 δ².

17. Circle on BC containing angle λ

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

a cos λ X − b cos (c + λ) Y − c cos (B + λ) Z = abc cos (A − λ).

18. Polar circles of the triangles BOC, COA, AOB, where O is the orthocentre 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.]

X = 2bc cos A, Y = 2ca cos B, Z = 2ab cos C.

The tripolar equation of the Nine-point circle may be obtained in the following manner.

Let X', Y', Z' be the squares of the distances of any point in the plane of the triangle ABC from the mid-points of the sides of that triangle.

The Nine-point circle being the circumcircle of the medial triangle, equation is

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

a/2 cos A X' +b/2 cos B Y' + c/2 cos C Z' = abc/8

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

Also Y+Z=2X'+a²/2, Z+X=2Y'+b²/2, and X +Y =2Z'+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.]

Hence σ {a cos A (Y+Z)} =½ {abc + σ(a³ cos a)}

from which we obtain

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

σ {a cos (B−C) X} = abc (1+2 cos A cos b cos C)

If two circles of radii R + ρ and R − ρ be described concentric with the circumcircle of the triangle ABC, the tripolar equation of the pair of circles is

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

a√X−ρ²+b√Y−ρ²+c√Z−ρ² = 0.

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

To prove this, let any point P be taken on the circumcircle: If λ + μ + ν = o, the co-ordinates of P may be taken to be 2δ/h(a/λ, b/μ, c/ν) where h = a²/λ+b²/μ+c²/ν.

The equation of the circle of radius ρ having its centre at P will be

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

a²/λ X + b²/μ Y + c²/ν Z = ρ² (a²/λ + b²/μ + c²/ν)

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

i.e., a²/λ (X − ρ²) + b²/μ (Y − ρ²) + c²/ν (Z − ρ²) = 0

and the envelope of this circle as λ, μ, ν vary is

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

a√X−ρ² + b√Y−ρ² + c√Z−ρ² = 0 (5)

and this envelope consists of two circles of radii R + ρ and R − ρ concentric with the circumcircle.

This equation (5) when expanded is

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

σ (a⁴X²) − 2σ (b²c²YZ) + 4abc ρ² σ (a cos A X) − 16 σ²ρ⁴ =0.

Subtracting this from the fundamental relation (4) multiplied by 4R² we have

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

4R²{σ (a² cos² AX²) + 2σ (bc cos B cos C YZ} − abc (2R²+ρ²) σ (a cos A X) + a²b²c²R² + 4δ²ρ⁴ = 0

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

i.e., {σ (a cos A X)}² − 2θ (2R²+ρ²) σ (a cos A X)+ θ² (4R⁴+ρ⁴) = 0

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

hence {σ (a cos A X) − θ (2R²+ρ²)}² = 4R²ρ²θ²,

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

or σ (a cos A X) = θ (R²+(R±ρ)²],

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

giving a cos A X + b cos B Y + c cos C Z = 4R sin A sin B sin C [R² + (R ± ρ)²],

a result in accord with that previously given.

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

In a similar manner by supposing the centre of a circle of radius ρ to move on the line la+mβ+nγ = 0 we obtain the equation of the pair of straight lines parallel to la+mβ+nγ = 0 and distant ρ from it in the form

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

l²/a²(F²−4YZ)+m²/b²(G²−4ZX)+n²/c²(H²−4XY) − 2mn/bc (GH-FX) − 2nl/ca(HF-GY) − 2lm/ab(FG-HZ)=0,

where

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

F≡Y+Z−a²−2ρ² G≡Z+X−b²−2ρ² H≡X+Y−c²−2ρ²

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

Circles of fixed radius r, whose centres are at the extremities of chords of the circle ABC passing through the symmedian point of that triangle, intersect on a circle whose centre is at the symmedian point, and whose radius is √r²−R² tan²ω, where ω is the Brocard angle 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.]

Consider the two points P, Q, whose trilinear ratios are (a/λ, b/μ, c/v') and (a/λ', b/μ', c/v') respectively where λ + μ + v = o, and λ' = μ—v, μ' = v—λ, v' = λ—μ. The two points satisfy the equation of the circle ABC—viz., a/a + b/β + c/γ = o, and the equation of the chord PQ, λλ a/a + μμ' β/b +vv' γ/c = o is satisfied by the co-ordinates of the symmedian point (a, b, c).

Hence the tripolar equations of the two circles of radius r whose centres are at P, Q are

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

a²/λ(X−r²)+b²/μ(Y−r²)+c²/ν(Z−r²) =0 a²/λ'(X−r²)+b²/μ'(Y−r²)+c²/ν'(Z−r²) =0

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

Hence a²(X−r²):b²(Y−r²):c²(Z−r²) = 1/μν' − 1/μ'ν: 1/νλ − 1/ν'λ: 1/λμ' − 1/λ'μ = λλ': μμ': νν'.

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

Therefore a²(X−r²)+b²(Y−r²)+c²(Z−r²) =0

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

or a²X+b²Y+c²Z = r²σ(a²),

which is the equation of a circle having its centre at the symmedian point of the triangle ABC.

The equation of a circle of radius r' having its centre at the symmedian point is

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

κσ(a²X)=κ²6Rabc+2δr'²

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

where κσ(a²)=2δ

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

Hence σ(a²r²=κ6Rabc + 2δr'²/κ =48δ²R²/σ9a² + σ (a²)r'²

Writing for σ (a²) its value 4δ cot ω, where ω is the Brocard angle, we have

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

r²—r'² = 3R² tan²ω

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

i.e., r' = √r²—3R² tan²ω

It follows from the above that the length of the minimum chord through the symmedian point is 2 √3 R tan ω.