form the equations of the polars, with respect to S+ P', of the three centres of similitude as above, we get (k'k'' — R) P′ = P", (k'k'"' — R′) P' = P'"', &c., showing that the line we want to construct is got by joining one of the four radical centres to the pole, with respect to S+ P2, of one of the four axes of similitude. This may also be derived geometrically as in Art. 121, from the theorems proved, Art. 306. The sixteen lines which can be so drawn meet +P in the thirty-two points of contact of the different conics which can be drawn to fulfil the conditions of the problem.* The solution here given is the same in substance (though somewhat simplified in the details) as that given by Prof. Cayley, Crelle, vol. XXXIX. Prof. Casey (Proceedings of the Royal Irish Academy, 1866) has arrived at another solution from considerations of spherical geometry. He shows by the method used, Art. 121 (a), that the same relation which connects the common tangents of four circles touched by the same fifth connects also the sines of the halves of the common tangents of four such circles on a sphere; and hence, as in Art. 121 (6), that if the equations of three circles on a sphere (see Geometry of Three Dimensions, chap. IX.) be S-L2 = 0, S - M2 = 0, S- N2 = 0, that of a group of circles touching all three will be of the form J{x (sa — L)} + J{μ (S* − M)} + √{v (S* − N)} = 0. This evidently gives a solution of the problem in the text, which I have arrived at directly by the following process. Let the conic S be x2+ y2+ 22, and let L = lx+my+nz, M = l'x + m'y + n'z; then the condition that S - L2, S-M2 should touch is (Art. 387) (1− S′) (1 − S′′) = (1 − K)2, where S′ = l2 + m2 + u2, S"=l'2+m'2+n'2, R=ll'+mm'+nn'. I write now (12) to denote √(1-S′)(1—S′′) − (1 − R). Let us now, according to the rule of multiplication of determinants, form a determinant from the two matrices containing five columns and six rows each. The resulting determinant which must vanish, since there are more rows than columns, is 0, 1, 1, 1, 1, 1 J(1 - S'), 0, (12), (13), (14), (15) = 0, an identical relation connecting the invariants of five conics all having double contact with the same conic S. Suppose now that the conic (5) touches the other four, Ex. 2. The four conics having double contact with a given one S, which can be drawn through three fixed points, are all touched by four other conics also having double contact with S.* Let S = x2 + y2 + z2 - 2yz cos A - 2zx cos B-2xy cos C, then the four conics are S = (x + y + 2)2, which are all touched by and by the three others got by changing the sign of A, B, or C, in this equation. Ex. 3. The four conics which touch x, y, z, and have double contact with S are all touched by four other conics having double contact with S. Let M = } (A+B+C'), then the four conics are S= {x sin (M- A) + y sin (M – B) + z sin (M — C')}2, together with those obtained by changing the sign of A, B, or C in the above; and one of the touching conics is the others being got by changing the sign of x, and at the same time increasing B and C by 180°, &c. Ex. 4. Find the condition that three conics U, V, W shall all have double contact with the same conic. The condition, as may be easily seen, is got by eliminating λ, μ, v between Δλ3 – Θλάμ + θ'λμ? - Δ' μ3 = 0, - and the two corresponding equations which express that μVvW, vW – XU break up into right lines. then (15), &c. vanish; and we learn that the invariants of four conics all having double contact with S and touched by the same fifth are connected by the relation or 0, (12), (13), (14) (12), 0, (23), (24) (13), (23), 0, (34) (14), (24), (34), 0=0, J{(12) (34)} ± J{(13) (24)} ± J{(14) (23)} = 0. We may deduce from this equation as follows the equation of the conic touching three others. If the discriminant of a conic vanish, S1, and then the condition of contact with any other reduces to R = 1. If, then, a, ẞ, y be the coordinates of any point satisfying the relation S- L2 = 0, or x2 + y2 + z2 (lx+my+nz)2 = 0, then - evidently denotes a conic whose discriminant vanishes and which touches S— L3. If, then, we are given three conics S-L2, S- M2, S- N2, take any point a, ẞ, y on the conic which touches all three and take for a fourth conic that whose equation has just been written, then the functions (14), (24), (34) are respectively J[(23) {J(S) — L}] ± √[(31) {√(S) − M}] ± √[(12) {√(S) — N}] = 0. *This is an extension of Feuerbach's theorem (p. 127), and itself admits of further extension. See Quarterly Journal of Mathematics, vol. VI. p. 67. 388. The theory of invariants and covariants of a system of three conics cannot be fully explained without assuming some knowledge of the theory of curves of the third degree. Given three conics U, V, W, the locus of a point whose polars with respect to the three meet in a point is a curve of the third degree, which we call the Jacobian of the three conics. For we have to eliminate x, y, z between the equations of the three polars U1x + U2y + U ̧z=0, V1x+ V ̧y+ V ̧2 = 0, W1x+ W ̧y + W ̧2=0, and we obtain the determinant U1(V ̧W ̧− V ̧W2) + U ̧ (V ̧W, − V ̧ W ̧) + U ̧(V1W ̧− V ̧W1) = 0. 8 2 3 1 2 It is evident that when the polars of any point with respect to U, V, W meet in a point, the polar with respect to all conics of the system lU+mV+n W will pass through the same point. If the polars with respect to all these conics of a point A on the Jacobian pass through a point B, then the line AB is cut harmonically by all the conics; and therefore the polar of B will also pass through A. The point B is, therefore, also on the Jacobian, and is said to correspond to A. The line AB is evidently cut by all the conics in an involution whose foci are the points A, B. Since the foci are the points in which two corresponding points of the involution coincide, it follows that if any conic of the system touch the line AB, it can only be in one of the points A, B; or that if any break up into two right lines intersecting on AB, the points of intersection must be either A or B, unless indeed the line AB be itself one of the two lines. It can be proved directly, that if lU+mV + nW represent two lines, their intersection lies on the Jacobian. For (Art. 292) it satisfies the three equations 1 lU1+ mV,+ n W1=0, lỤ ̧+mV ̧+nW ̧=0, lU ̧+mV ̧+nW ̧= 0; whence, eliminating l, m, n, we get the same locus as before. The line AB joining two corresponding points on the Jacobian meets that curve in a third point; and it follows from what has been said that AB is itself one of the pair of lines passing through that point, and included in the system IU+ mV + nW. The general equation of the Jacobian is (ag′h") x3 + (bh'ƒ'') y3 + (cf'g'') z3 — {(ab'g') + (ah'ƒ'')}x3y—{(ca′h")+(af′9′′)}x3z—{{(ab′ƒ'')+ (bg′h")}y*x —{(bc'h′′) + (bƒ'g'')}y'z—{(ca′ƒ'')+(cg'h'')}z2x— {(bc′g'') + (ch'ƒ'')} z3y - {(ab'd')+2(fg'h')} xyz=0, where (ag'h") &c. are abbreviations for determinants. Ex. 1. Through four points to draw a conic to touch a given conic W. Let the four points be the intersection of two conics U, V; and it is evident that the problem admits of six solutions. For if we substitute a+ ka', &c. for a in the condition (Art. 372) that U and W should touch each other, k, as is easily seen, enters into the result in the sixth degree. The Jacobian of U, V, W intersects W in the six points of contact sought. For the polar of the point of contact with regard to W being also its polar with regard to a conic of the form λU+μV passes through the intersection of the polars with regard to U and V. Ex. 2. If three conics have a common self-conjugate triangle, their Jacobian is three right lines. For it is verified at once that the Jacobian of ax2 + by2 + cz2, a'x2 + b'y2 + c'z2, a′′x2 + b′′y2 + c''z2 is xyz = 0. Ex. 3. If three conics have two points common, their Jacobian consists of a line and a conic through the two points. It is evident geometrically that any point on the line joining the two points fulfils the conditions of the problem, and the theorem can easily be verified analytically. In particular the Jacobian of a system of three circles is the circle cutting the three at right angles. Ex. 4. The Jacobian also breaks up into a line and conic if one of the quantities 8 be a perfect square L. For then L is a factor in the locus. Hence we can describe four conics touching a given conic S at two given points (S, L) and also touching S"; the intersection of the locus with S" determining the points of contact. When the three conics are a conic, a circle, and the square of the line at infinity, the Jacobian passes through the feet of the normals which can be drawn to the conic through the centre of the circle. 388 (a). We return now to the theory of two conics which it was not possible to complete until we had explained the nature of Jacobians. We have seen that a system of two conics S, S' has four invariants A, O, O', A', and a covariant conic F, but there is besides a cubic covariant. In fact, the covariant conic F has a common self-conjugate triangle with S, S' (Art. 381, Ex. 1), therefore (Art. 388, Ex. 2) if we form J the Jacobian of S, S', F we obtain a cubic covariant, which, in fact, represents the sides of the common self-conjugate triangle of S end S'. It appears from (Art, 378a) that J vanishes identically if S and S' have double contact. We have given (Art, 381, Ex. 4) another method of obtaining the equation of the sides A A A. of the common self-conjugate triangle, and if we compare the results of the two methods, we get the identical equation J2 = F3 – F2 (©S′ + O ́S) + F (A′OS2 + AO'S′′2) +FSS' (O'-3▲▲′) – A′′AS® - A3A'S" + A′ (2AO′ – O2) S2 S′ + ▲ (2A′O – O'2) SS”. Thus we see that a system of two conics has, besides the four invariants, four covariant forms S, S', F, J, these being connected by the relation just written. In like manner, there are four contravariant forms 2, 2, 4, г, where the last expresses tangentially the three vertices of the self-conjugate triangle, its square being connected by a relation, corresponding to that just written, between 2, ', and the invariants. Ex. 1. Write down the 12 forms for the conics x2 + y2 + z2, ax2 + by2 + cz2. S = x2 + y2 + z2, S′ = ax2 + by2+cz2, F = a (b+c) x2 + b (c + a) y2 + c (a+b) z2, Σ = λ2 + μ2 + v2, Σ' = bcλ2 +caμ2+abv2, • = (b+c) X2 + (c+a) μ2 + (a+b) v2, Ex. 2. Find an expression for the area of the common conjugate triangle of two conics. The square of the area is found to be M2 sin A sin2B sin2C where M is the area of the triangle of reference, and r' the result of substituting in I, sin A, sin B, sin C, the coordinates of the line at infinity. That the expression must contain in the numerator the condition of contact, and in the denominator I", is evident from the consideration that this area must vanish if the conics touch, and becomes infinite if any vertex of the triangle be at infinity. 388 (b). We have already explained what is meant by covariants which express relations satisfied by x, y, z, the coordinates of a point lying on a locus having some permanent relation with the original curve or curves, and by contravariants which express relations satisfied by λ, u, the tangential coordinates of a line, whose section by the original curve or curves has some property unaffected by transformation of coordinates. There are besides forms called mixed concomitants which contain both x, y, z and also λ, u, v, and these we proceed |