COR. It appears, from the preceding investigation, that if a, B, y be the co-ordinates of the centre of the conic represented by the equation γ $ (a, B, y) = ua2 + vß2 + wy2 + 2u'By + 2v'ya + 2w'aß =0, 13. To find the condition that the conic may be a rectangular hyperbola. If the equations of the asymptotes be la + mB+ny = 0, l'a+m'B+n'y = 0, the condition of their perpendicularity is ll' + mm' + nn' — (mn' + m'n) cos A – (nl' +n'l) cos B - Ua2 + Vb2 + Wc2 + 2 U'bc + 2 V'ca + 2 W'ab = D, 12 uvw + 2u'v'w' — uu'2 — vv12 — ww12 = K, we see, by reference to Art. 12, that nn' Dv – Kb2 ̄ Dw – Kc2 Du-Ka2 + (nl' + n'l) † (lm' + l'm) = Dw'-Kab Hence the required condition is D(u+v+w−2u' cos A – 2v′ cos B-2w cos C) − K (a2 + b2 + c2 −2bc cos A-2ca cos B-2ab cos C)=0. 2 Now a+b2+ c2 — 2bc cos A-2ca cos B – 2ab cos C = 0 identically, hence the required condition becomes u+v+w-Qu' cos A-2v' cos B-2w' cos C= 0. COR. It hence appears, that the condition that the conic u'By + v'ya + w'aß = 0, described about the triangle of reference, may be a rectangular hyperbola, is ucos A+ v' cos B+ w' cos C=0; that is, the conic must pass through the point determined by the equations a cos A = ẞ cos B = y cos C. This point (see Art. 5, Chap. 1.) is the point of intersection of the perpendiculars let fall from each angular point of the triangle on the opposite side. Hence we obtain the following elegant geometrical proposition, that every rectangular hyperbola described about a given triangle passes through the point of intersection of the perpendiculars let fall from each angular point of the triangle on the opposite side. Again, if u', v', w' be all = 0, the condition is u+v+w=0, proving that, if the equation ua2 + vß2 + wy2 = 0 represent a rectangular hyperbola, the curve will pass through the four points for which a=B=y, a=B = y, a = -ß=y, a=ß=—7. CONDITIONS FOR A CIRCLE. 85 In other words, if a rectangular hyperbola be so described that each angular point of a given triangle is the pole, with respect to it, of the opposite side, it will pass through the centres of the four circles which touch the three sides of the triangle. 14. To investigate the conditions that the general equation of the second degree shall represent a circle. The property of the circle, which we shall assume as the basis of our investigation, is the following: that if, through any point, chords be drawn cutting a circle, the rectangle, contained by their segments, is invariable. tion cut Suppose then, that the curve, represented by the equa ua2 + vß2 + wy2+ 2u'By + 2v'ya +2w'aß = 0, Let h, h' be the respective distances of c2, a, from AB; g, g', those of a,, b, from AC: then, multiplying the first of the above three equations by sin' A, we get hh' =gg'. Now h, h' are the two values of y obtained by putting B0 in the equation of the conic section, bearing in mind that, when ẞ=0, = aa+cy = 24. This gives, for the determination of y, the equation u (cy — 2A)2 + wa3y2 + 2av'y (2▲ — cy) = 0 ; whence, by the theory of equations, Hence, since Ac,. Aa,= Aa ̧ . Ab,, we obtain uc2 + wa2 - 2v'ca = va2 + ub2 — 2w'ab. gives wb2+vc2-2u'bcuc2+wa-2v'ca, which also follows from the preceding two equations. Hence the equations wb2+vc2 — Qu'bc = uc2 + wa2 - 2v'ca = va2+ub2 - 2w'ab are necessary conditions that the given equation should represent a circle; and, since they are two in number, they are sufficient. INTERSECTION WITH THE LINE AT INFINITY. 87 15. To determine the intersection of a circle with the line at infinity. Since, at every point in the line at infinity, Substituting these values in the equation (2u'bc — vc2 — wb2) aẞy + (2v'ca — wa2 — uc2) bya which, if the conic be a circle, reduces to aßy+bya + caß = 0, shewing that every circle intersects the line at infinity in the same two points as the circle described about the triangle of reference; that is, all circles intersect the line at infinity in the same two points. These points are, of course, imaginary. From this it follows that every circle may be represented in either of the following forms, aßy+bya + caß + (la+mß +ny) (aa +bB+ cy) = 0, sin 2A.a2+sin 2B.ß2+sin2 C.y2+(λa+μß+vy) (aa+bB+cy)=0. |