the sides of the quadrilateral. The line joining the point of intersection of two of these lines with the point of intersection of the other two is called a diagonal of the quadrilateral. There are therefore three diagonals, viz. PQ, AC, BD in the figure. We must take the negative sign, for two of the rays coincide if the anharmonic ratio of a pencil be equal to +1. P"Q" This follows from Art. 55, for if =1,. then P" and R" R"Q" are coincident. Hence the diagonal AC is cut harmonically. We can prove in a similar manner that the other diagonals are divided harmonically. INVOLUTION. *61. Def. Let O be a fixed point on a given straight line, and P, P'; Q, Q'; R, R'; &c. pairs of points on the line such that OP. OP=OQ.OQ' OR. OR' = ...... a const. k. = Then these points are said to form a system in involution, of which the point 0 is called the centre. Two points. such as P, P ́ are said to be conjugate to one another. The point conjugate to the centre is at an infinite distance. If each point be on the same side of the centre as its conjugate, there will be two points K1, K2, one on each side of the centre, such that OK, OK, OP. OP'. These points K1, K, are called double points or foci. 2 2 = It is clear that when the two foci are given the involution is completely determined. An involution is also completely determined when two pairs of conjugate points are given. For, let a, a and b, b' be the distances of these points from any point in the straight line upon which they lie, and let a be the distance of the centre of the involution from that point. Then we have the relation or (α − x) (a' — x) = (b − x) (b′ — x), Hence there is only one position of the centre. The position of the centre can be found geometrically by drawing circles one through each of the two pairs of conjugate points, then [Euclid III. 37] the common chord of the circles will cut the line on which the points lie in the required centre. *62. If any number of points be in involution the cross ratio of any four points is equal to that of their four conjugates. Let P, Q, R, S be any four points, and let the distances of these points from the centre be p, q, r, s respectively and kkk k therefore those of their conjugates P'q'respectively. (q-p) (s-r) (s− p) (q − r) ' kkkk {PQRS} = p S r/ (p − q) (r− s) S p k r = (p − s) (r− q). {PQRS} = {P'Q'R'S'}. The above gives us at once a means of testing whether or not six points are in involution. For P, P' will be con jugate points in the involution determined by A, A′ and B, B', if {ABA'P}={A'B'AP'}. *63. Any two conjugate points of an involution and the two foci form a harmonic range. 2 Let K, K, be the two foci, and O the centre of the involution, and let K,O=c= OK. that or or Then, if P, P′ be the two conjugates we have to prove (c + OP)(OP' − c) + (c+ OP')(OP — c) = 0, OP. OP' = c2, which we know to be true. *64. If any number of pairs of points in involution be joined to any point O we obtain a pencil of lines which may be said to be in involution. Such a pencil is cut by any other transversal in pairs of points which are in involution. This follows from Articles 55 and 62. EXAMPLES. 1. If P, Q, R, S be any four points on a straight line, then PQ.RS+PR.SQ+PS.QR=0. 4. Shew that, by taking four points in different orders, six and only six different cross ratios are obtained, and that of these six three are the reciprocals of the other three. 5. If {PQRS}=-1, shew that {SRQP}=-1, {PRQS}=2, and {PSQR}=}. 6. If {PQRS} = − 1, and O be the middle point of PR, then CHAPTER IV. THE CIRCLE. 65. To find the equation of a circle referred to any rectangular axes. Let C be the centre of the circle, and P any point on its circumference. Let d, e be the co-ordinates of C'; x, y the co-ordinates of P; and let a be the radius of the circle. Draw CM, PN parallel to OY, and CK parallel to OX, as in the figure. Then is the required equation. (i), If the centre of the circle be the origin, d and e will beth be zero, and the equation of the circle will be x2 + y2 = a2.. The equation (i) may be written (ii). x2 + y2 - 2dx-2ey + d2 + e2 — a2 = 0. The equation of any circle is therefore of the form. x2 + y2+2gx+2fy + c = 0.........(iii), where g, f and c are constants. Conversely the equation (iii) is the equation of a circle. For it may be written (x+g)2 + (y +ƒ)2 = g3 +ƒa − c ; and this last equation shews that the distance from any point on the locus of the equation (iii) from the point (-9,-f) is constant and equal to g2 +f2 — c. The equation (iii) therefore represents a circle of radius g+f-c, the centre of the circle being at the point -9,-f). 1 If g2+f2-c=0 the radius of the circle is zero, and the circle is called a point-circle. y If g2+f2- c be negative, no real values of x and of will satisfy the equation, and the circle is called an imaginary circle. 66. We have seen that the general equation of a circle is x2 + y2+2gx+2fy + c = 0. This equation contains three constants. If we want to find the equation of a circle which passes through three given points, or which is defined in some other manner, we assume the equation to be of the above form and determine the values of the constants g, f, c for the circle in question from the given conditions. |