15.1 The power of a point with respect to a circle. , so that P = xa+(y +z)x. Applying Stewart s theorem in succession to triangles QAX, QBC.

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1 Chapter 15 Circle equations 15.1 The power of a point with respect to a circle The power of P with respect to a circle Q(ρ) is the quantity P(P) := PQ 2 ρ 2. A point P is in, on, or outside the circle according as P is negative, zero, or positive. Proposition Let P = xa + yb + zc in absolute barycentric coordinates. P(P) = xp(a)+yp(b)+zp(c) (a 2 yz +b 2 zx+c 2 xy). A P Q B Proof. Let X be the trace of P on BC. In absolute coordinates, X = yb+zc y+z, so that P = xa+(y +z)x. Applying Stewart s theorem in succession to triangles QAX, QBC andabc, we have X C PQ 2 = xaq 2 +(y +z)xq 2 x(y +z)ax 2 = xaq 2 +(y +z) ( y y +z BQ2 + z y +z CQ2 = xaq 2 +ybq 2 +zcq 2 yz y +z BC2 x(y +z)ax 2 = xaq 2 +ybq 2 +zcq 2 yz y +z BC2 ( z x(y +z) y +z AC2 + y y +z AB2 yz ) (y +z) 2 BC2 yz ) (y +z) 2BC2 x(y +z)ax 2 = xaq 2 +ybq 2 +zcq 2 (1 x)yz BC 2 zx AC 2 xy AB 2 y +z = xaq 2 +ybq 2 +zcq 2 (a 2 yz +b 2 zx+c 2 xy).

2 422 Circle equations From this it follows that P(P) = PQ 2 ρ 2 = x(aq 2 ρ 2 )+y(bq 2 ρ 2 )+z(cq 2 ρ 2 ) (a 2 yz +b 2 zx+c 2 xy) = xp(a)+yp(b)+zp(c) a 2 yz b 2 zx c 2 xy Circle equation Using homogeneous barycentric coordinates for P and writing f := P(A), g := P(B), h := P(C), for the powers ofa,b, C with respect to a circle Q(ρ), we have, P(P) = fx+gy +hz x+y +z a2 yz +b 2 zx+c 2 xy (x+y +z) 2 = (a2 yz +b 2 zx+c 2 xy) (x+y +z)(fx+gy +hz) (x+y +z) 2. Therefore, the equation of the circle is (a 2 yz +b 2 zx+c 2 xy) (x+y +z)(fx+gy +hz) = 0. Example (1) The equation of the circumcircle is a 2 yz + b 2 zx + c 2 xy = 0 since f = g = h = 0. (2) For the incircle, we have The equation of the incircle is f = (s a) 2, g = (s b) 2, h = (s c) 2. a 2 yz +b 2 zx+c 2 xy (x+y +z)((s a) 2 x+(s b) 2 y +(s c) 2 z) = 0. (3) Similarly, the A-excircle has equation a 2 yz +b 2 zx+c 2 xy (x+y +z)(s 2 x+(s c) 2 y +(s b) 2 z) = 0. (4) For the nine-point circle, we have f = b 2 ccosa = b 2 Sα b = 1 2 S α. Similarly, g = 1 2 S β andh = 1 2 S γ. Therefore, the equation of the nine-point circle is 2(a 2 yz +b 2 zx+c 2 xy) (x+y +z)(s α x+s β y +S γ z) = 0.

3 15.3 Points on the circumcircle 423 Exercise 1. Find the equation of the Conway circle. 2. Find the equations of the circles (i) C BBC passing through B andc, and tangent to BC atb, (ii) C BCC passing through B andc, and tangent tobc atc. 3. Compute the coordinates of the Brocard points: (i) Ω as the intersection of the circles C BBC,C CCA, andc AAB, (ii) Ω as the intersection of the circles C BCC,C CAA, andc ABB. 4. Find the equation of the circle with diameter BC Points on the circumcircle The equation of the circumcircle can be written in the form a 2 x + b2 y + c2 z = 0. This shows that the circumcircle consists of the isogonal conjugates of infinite points X(101) The point is clearly on the circumcircle. ( ) a 2 X(101) = b c : b 2 c a : c 2 a b X(100) The point X(100) = ( ) a b c : b c a : c a b is clearly on the circumcircle. It is the isogonal conjugate of the infinite point (a(b c) : b(c a) : c(a b)) (on the trilinear polar of the incenter, namely, the line x a + y b + z c = 0). Its inferior is a point on the nine-point circle. To find this, we rewrite X(100) = (a(c a)(a b) : b(a b)(b c) : c(b c)(c a)).

4 424 Circle equations From this, the Feuerbach point! inf(x(100)) = (b(a b)(b c)+c(b c)(c a) : : ) = ((b c)(b(a b)+c(c a)) : : ) = ((b c) 2 (b+c a) : : ), 1. The distance from X(100) to the Nagel point is the diameter of the incircle. 2. X(100) is the intersection of the Euler lines of the triangles I a BC,I b CA,I c AB The Steiner point X(99) The Steiner point X(99) = The inferior of the Steiner point is ( ) 1 b 2 c : 1 2 c 2 a : 1 2 a 2 b 2 X(115) = ((b 2 c 2 ) 2 : (c 2 a 2 ) 2 : (a 2 b 2 ) 2 ). This is the midpoint between the Fermat points The Euler reflection point E = X(110) Theorem The reflections of the Euler line in the sidelines of triangle ABC are concurrent at ( ) a 2 E = b 2 c : b 2 2 c 2 a : c 2 2 a 2 b 2 on the circumcircle. Proof. The Euler line intersects the sideline BC at S α (S β S γ )x+s β (S γ S α )y +S γ (S α S β )z = 0 X = (0 : S γ (S α S β ) : S β (S γ S α ). We find the reflection of H in BC as follows. From the relation (S βγ +S γα +S αβ )(0,S γ,s β )+S βγ (S β +S γ, S γ, S β ) = (S β +S γ )(S βγ,s γα,s αβ ), we have H = X + S βγ (S β +S γ )(S βγ +S γα +S αβ ) (S β +S γ, S γ, S β ).

5 15.4 Circumcevian triangle 425 Therefore, its reflection inbc is the point X = X In homogeneous barycentric coordinates, this is S βγ (S β +S γ )(S βγ +S γα +S αβ ) (S β +S γ, S γ, S β ). X = (S βγ +S γα +S αβ )(0,S γ,s β ) S βγ (S β +S γ, S γ, S β ) = ( S βγ (S β +S γ ),S γ (2S βγ +S γα +S αβ ),S β (2S βγ +S γα +S αβ )) The inferior of the Euler reflection is the point X(125) = ((b 2 c 2 ) 2 (b 2 +c 2 a 2 ) : (c 2 a 2 ) 2 (c 2 +a 2 b 2 ) : (a 2 b 2 ) 2 (a 2 +b 2 c 2 )). This is the intersection of the Euler lines of the trianglesayz,bzx,cxy, wherexyz is the orthic triangle Circumcevian triangle Let P = (u : v : w). The lines AP, BP, CP intersect the circumcircle again at X, Y, Z. The triangle XYZ is called the circumcevian triangle of P. Since X lies on the line AP, X = (x : v : w) for some x. This point lies on the circumcircle if and only if This gives x = a2 vw b 2 w+c 2 v. Therefore, Similarly, a 2 vw+b 2 xw +c 2 xv = 0. X = ( a 2 vw : (b 2 w +c 2 v)v : (b 2 w +c 2 v)w). Y = ((c 2 u+a 2 w)u : b 2 wu : (c 2 u+a 2 w)w), Z = ((a 2 v+b 2 u)u : (a 2 v+b 2 u)v : c 2 uv). Proposition The circumcevian triangle of P = (u : v : w) is perspective with the tangential triangle at ) ) (a ( 2 a4 u + b4 2 v + c4 : :. 2 w 2 Proof. The vertices of the tangential triangle are ( a 2,b 2,c 2 ), (a 2, b 2,c 2 ), (a 2,b 2, c 2 ). The line joining ( a 2,b 2,c 2 ) tox is x y z a 2 b 2 c 2 a 2 vw (b 2 w+c 2 v)v (b 2 w +c 2 v)w = 0.

6 426 Circle equations This is (b 4 w 2 c 4 v 2 )x+a 2 b 2 w 2 y a 2 c 2 v 2 z = 0. Similarly, the lines joining (a 2, b 2,c 2 ) toy and(a 2,b 2, c 2 ) toz are a 2 b 2 w 2 x+(c 4 u 2 a 4 w 2 )y +b 2 c 2 u 2 z = 0, a 2 b 2 v 2 x b 2 c 2 u 2 y +(b 4 v 2 b 4 u 2 )z = 0. These three lines concur at a point with coordinates given above. Example G: X(22) = (a 2 ( a 4 +b 4 +c 4 ) : : ). ( ) 2. H: X(24) = a 2 (a 4 +b 4 +c 4 2a 2 b 2 2a 2 c 2 ) : :. b 2 +c 2 a 2 These two points are on the Euler line, and are the centers of similitude of the circumcircle and incircle of the tangential triangle The third Lemoine circle Given a pointp = (u : v : w), it is easy to find the equation of the circlec a throughp,b, C. Since P(B) = P(C) = 0, the equation of the circle is C a : a 2 yz +b 2 zx+c 2 xy (x+y +z) fx = 0 for some f. Since the circle passes through P = (u : v : w), we must have f = a2 vw+b 2 wu+c 2 uv. u(u+v +w) This circle C a intersects the lines AC and AB each again at another point. To find the intersection withac, we puty = 0 in the equation of(c a ) and obtainb 2 zx fx(x+z) = 0, x((b 2 f)z fx)) = 0. Therefore, apart fromc = (0,0,1), the circlec a intersectsac at B a = (b 2 f : 0 : f) = (b 2 u 2 +b 2 uv a 2 vw c 2 uv : 0 : a 2 vw+b 2 wu+c 2 uv). Similarly, the circle C a intersects AB again at C a = (c 2 f : f : 0) = (c 2 u 2 +c 2 wu a 2 vw b 2 wu : a 2 vw+b 2 wu+c 2 uv : 0). Similarly, with g = a2 vw+b 2 wu+c 2 uv v(u+v +w) and h = a2 vw+b 2 wu+c 2 uv, w(u+v +w) the circles C b through P, C,AandC c through P,A,B intersect the sidelines again at C b = (g : c 2 g : 0), A b = (0 : a 2 g : g), A c = (0 : h : a 2 h), B c = (h : 0 : b 2 h).

7 15.5 The third Lemoine circle 427 and Note the lengths of the segments: AB a = f b 2 b = f b, AC a = f c 2 c = f c, AB c = b2 h b 2 AC b = c2 g The four points B a,b c,c a,c b are concyclic if and only if c 2 b = b2 h, b c = c2 g. c AB a AB c = AC a AC b = f(b2 h) b 2 = f(c2 g) c 2 = b2 c 2 = h g = v2 w 2. Likewise, the four points C b, C a, A b, A c are concyclic if and only if c2 = w, and the a 2 u four points A c,a b,b c,b a are concyclic if and only if a2 = u. b 2 v By the principle of 6 concyclic points, the six points A b, A c, B c, B a, C a, C b are concyclic if and only if u : v : w = a 2 : b 2 : c 2, namely, P = (u : v : w) = (a 2 : b 2 : c 2 ), the symmedian point. The circle C containing these 6 points is the third Lemoine circle. For this choice ofp, f = 3b 2 c 2 a 2 +b 2 +c 2, g = 3c 2 a 2 a 2 +b 2 +c 2, h = 3a 2 b 2 With respect to the circle C containing these 6 points, we have Similarly, a 2 +b 2 +c 2. P(A) = f(b2 h) b 2 = 3b2 c 2 b 2 (b 2 +c 2 2a 2 ) b 2 (a 2 +b 2 +c 2 ) 2 = 3b2 c 2 (b 2 +c 2 2a 2 ) (a 2 +b 2 +c 2 ) 2. P(B) = 3c2 a 2 (c 2 +a 2 2b 2 ) (a 2 +b 2 +c 2 ) 2, P(C) = 3a2 b 2 (a 2 +b 2 2c 2 ) (a 2 +b 2 +c 2 ) 2. From these, we obtain the equation of the third Lemoine circle: (a 2 +b 2 +c 2 ) 2 (a 2 yz +b 2 zx+c 2 xy) 3(x+y +z) ( b 2 c 2 (b 2 +c 2 2a 2 )x+c 2 a 2 (c 2 +a 2 2b 2 )y +a 2 b 2 (a 2 +b 2 2c 2 )z ) = 0.