And the polar of (f, g, h) is fa+gB+ hy = 0.

Hence if the line fa+gB+hy =0 touch (1), the point (f, g, h) lies in (2), giving for the condition of tangency

And if the line fa+gB+hy=0 touch (2), the point (f, g, h) lies in (1), giving for the condition of tangency in that case

X2ƒ2 + μ3g2 + v3h2 — 2μvgh — 2vλhƒ — 2λμfg = 0.

These conditions of tangency are identical with those already investigated.

Again, every parabola touches the line at infinity. Now the co-ordinates of the pole of this line are proportional to a, b, c. Hence, if the conic, represented by the general equation of the second degree, be a parabola, the point (a, b, c) must lie in the reciprocal conic. This gives, as the condition for a parabola,

Ua2+ Vb2 + We2 + 2U'bc + 2 V'ca + 2W'ab= 0,

the same as that already investigated.

15. PROP. Any straight line drawn through a given plane A is divided harmonically by any conic section, and the polar of A with respect to it.

This proposition may be proved as follows. Let the straight line cut the curve in P and Q, and the polar of A in B. Let C be the polar of the straight line, and let ABC be the triangle of reference. The conic will be self-conjugate with respect to ABC, and will therefore be represented by the equation

ua2 + vß2 + wy2 = 0.

Hence the lines CP, CQ, which are tangents to the conic, are represented by the equation

ua2 + B2 = 0,

and therefore form an harmonic pencil with CA, CB.

16. The straight line CB may be regarded as the polar of A with respect to the locus made up of the two straight lines CP, CQ. For the values of B and y at A being 0, 0, and the equation of CP, CQ being ux2+vß2 = 0, we get for the equation of its polar, a = 0, that is, the polar is the line AB,

17. If four straight lines form an harmonic pencil, either pair will be its own polar reciprocal with respect to the other. For, adapting the equation of Art. 8, to the case of two variables only, we get for the polar reciprocal of aß = 0, with respect to uz2+vß*= 0, the following equation,

And, conversely, for that of ua2 + vß2 = 0 with respect to aẞ=0,

or ux2+vß2 = 0, in either case reproducing the reciprocated Hence the proposition is proved.

curve.

18. We may hence deduce the condition that two pairs of straight lines may form an harmonic pencil. First let them all intersect in A, and the equations of the two pairs be

ua2 + vß2 + 2w'aß = 0.............

pa2 +qß2+2r'aß = 0..........

The polar reciprocal of (1) with respect to (2) is

(1),

(2).

or ́`u (r'a+qß)2+v (pa + v′ ß)2 — 2w' (r'a+qß) (pa+r′ß) = 0.

Suppose that

ua2 + vß2 + 2w'aß = u (a + k‚ß) (a + k2ß)

identically, i. e. that

v = uk,k,,

2w' = u(k,+ k).

At the point of intersection of the line a+k,B=0, with y= 0, we have

Taking the polar of this with respect to the curve (2) we get

If this be identical with a + kß = 0, we get

The symmetry of this equation shews that (2) is also its own polar reciprocal with respect to (1), as ought to be the

case.

19. If the point of intersection of the four straight lines do not coincide with one of the angular points of the triangle of reference, we have then only to express the condition that the range formed by their intersection with any one of its sides, y= 0, for instance, be an harmonic range. If this be the case, the pencil formed by joining these four points with C will be an harmonic pencil, and we shall have, as before,

pv-2r'w' + qu= 0.

20. We next proceed to consider the results to be deduced from the theory of reciprocal polars, when the auxiliary conic is a circle. It is here that the utility of theory is most apparent, as we are thus enabled to transform metrical theorems, i.e. theorems relating to the magnitudes of lines and angles.

We know that, if PQ be the polar of a point T with respect to a circle, of which the centre is S and radius k, then ST will be perpendicular to PQ. Let ST cut PQ in V. Then

ST. SV=K.

Hence the pole of any line is at a distance from the centre of the auxiliary circle inversely proportional to the distance of the line. And conversely, the polar of any point is at a distance from the centre of the auxiliary circle, inversely proportional to the distance of the point itself.

21. If TX, TY be any two indefinite straight lines, P, Q their poles, then, since SP is perpendicular to TX, SQ to TY, it follows that the angle PSQ is equal to the angle XTY or its supplement, as the case may be. Hence, the angle included between any two straight lines is equal to the angle subtended at the centre of the auxiliary circle by the straight line joining their poles, or to its supplement.

22. From what has been said in Art. 15, and the earlier articles of this chapter, it will appear that to find the polar reciprocal of a given curve with respect to a circle, we may proceed by either of the following two methods.

First. Draw a tangent to the curve, and from S, the centre of the auxiliary circle, draw SY perpendicular to the tangent, and on SY, produced if necessary, take a point Q, such that SQ. SY=k. The locus of Q will be the required polar reciprocal.

Secondly. Take a point P on the curve, and join SP; on SP, produced if necessary, take a point Z, such that

SP. SZ=k2.

Through Z draw a straight line perpendicular to SP. The envelope of this line will be the required polar reciprocal.

23. It will be observed that the magnitude of the radius of the auxiliary circle affects the absolute, but not the relative, magnitudes or positions of the various lines in the reciprocal figure. As our theorems are, for the most part, independent of absolute magnitude, we may generally drop all consideration of the radius of the auxiliary circle, and consider its centre only. We may then speak of reciprocating "with respect to S" instead of "with respect to a circle of which Sis the centre." S may be called the centre of reciprocation, k the constant of reciprocation.

24. As an example of the power of this method we will reciprocate the following theorem, "The three perpendiculars from the angular points of a triangle intersect in a point."

This may be expressed as follows: "If O, A, B, C be four points, such that OB is perpendicular to CA, and OC to AB, then will OA be perpendicular to BC."

Reciprocate this with respect to any point S, and the four points O, A, B, C give four straight lines, which we may call each by three letters abc, ab'c', a'bc', a'b'c, respectively. Then, the fact that OB is perpendicular to CA is expressed by band b' subtending a right angle at S, or by bb' being a right angle. Again, the fact that OC is perpendicular to AB, shews that cSc' is a right angle. Then the reciprocal theorem tells us that a Sa' is also a right angle. We may express this more neatly as follows: aa', bb', cc', are the diagonals of the complete quadrilateral formed by the four straight lines, hence it appears that at any point at which two of the diagonals of a complete quadrilateral subtend a right angle, the third diagonal also subtends a right angle. Or, in other words, The three circles, described on the diagonals of a complete quadrilateral as diameters, have a common radical axis.

The extremities of this axis may be conveniently called the foci of the quadrilateral*.

25. If the system formed by the four points O, A, B, C be reciprocated with respect to any one of them, O for instance, the triangle thus obtained will be similar, and similarly situated, to that formed by the other three points A, B, C.

* This name is proposed by Mr Clifford, in the Messenger of Mathematics.