which may also be written.

a2 (p − q) (p − r) + b2 (q − r) (q − p) + c2 (r − p) (r − q) = 0, the equation of the two circular points at infinity.

Hence may also be deduced the conditions that the equation

up2 + vq2 + wr2 + 2u'qr + 2v'rp + 2w'pq=0

may represent a circle. For, comparing this equation with that just obtained (Art. 5, Cor.), we get

Putting each member of these equations = k, they may be

(b2 -vk) (c2-wk) = (bc cos A + u'k)2,

or (vw — u'2) k2 — (vc2 + wb2 + 2u'bc cos A) k+b2c2 sin2 A = 0.

Again, we get

(a-uk) (bc cos A+uk) + (ca cos B+ v'k) (ab cos C+w'k) = 0, or (v'w' — uu')k2+{a(au'+bcos C.v'+ccos B. w') – bc cos A.u}k + abc sin B sin C = 0.

b2c2

2

Now, since b2c2 sin2 A = a2bc sin B sin C = (2A), these equations may be written under the form

(vw — u ́2) k2 — (vc2 + wb2 + 2u'bc cos A) k + 4A2 = 0,

(v'w' — uu' ) k2+{a(au'+bv'cos C'+cw'cos B)—bc cos A. u}k+4A2=0.

Combining these with the four similar equations, we get

— {u (b2 + c2 — 2bc cos A) + va2 + wa2 + 2a3u'

+2av' (c cos B+b cos C) + 2aw' (b cos C+ c cos B)} = 0,

or

=

(u+v+w+2u' + 2v + 2w') a* k'

Two other corresponding expressions may of course be obtained for k, and the required condition is therefore

11. To find the equation of the conic with respect to which the triangle of reference is self-conjugate.

Since each angular point of the triangle of reference is the pole, with respect to this conic, of the opposite side, it follows that the equation of such a conic will be of the form

up2 + vq2 + wr2 = 0.

From the last article it will be seen that the equation of the self-conjugate circle is

1. A parabola is described about a triangle so that the tangent at one angular point is parallel to the opposite side; shew that the square roots of the perpendiculars on any tangent to the curve are in arithmetical progression.

2. A conic is circumscribed about a triangle such that the tangent at each angular point is parallel to the opposite side; shew that, if p, q, r be the perpendiculars from the angular points on any tangent,

3.

=

Shew that the equation of the centre of this conic is

p+q+r= = 0.

4. Conics are drawn each touching two sides of a triangle at the angular points and intersecting in a point; prove that the intersections of the tangents at this common point with the sides cutting their respective conics lie on one straight line, and that the common tangents to the conics intersect the sides in the same three points.

5. A system of hyperbolas is described about a given triangle; prove that, if one of the asymptotes always pass through a fixed point, the other will always touch a fixed conic, to which the three sides of the triangle are tangents.

6. A parabola touches one side of a triangle in its middle point, and the other two sides produced; prove that the perpendiculars, drawn from the angular points of the triangle upon any tangent to the parabola, are in harmonical progression.

7. Prove that the nine-point circle of the triangle of reference is represented by the equation

a (q + r)* + b (r + p)* + c (p + q)* = 0.

8. If p1, P2, P be the radii of three circles, the internal common tangents to each pair of the circles touch a conic whose tangential equation, referred to the centres of the circles, is

a2 (p − q) (p − r) + b2 (q − r) (q − p) + c2 (r − p) (r − q)

12. There is another system of Tangential Co-ordinates, which bears a close analogy to the ordinary Cartesian system. If x, y be the Cartesian co-ordinates of a point, referred to two rectangular axes, then the intercepts on these axes of the

polar of the point, with respect to a circle whose centre is the origin, and radius k, will be

k2 k2
XC y

respectively. These intercepts completely determine the position of the line, and their reciprocals may be taken as its co-ordinates, and denoted by the letters §, n.

13. In this system, every equation of the first degree represents a point.

be an equation of the first degree.

Draw the straight lines OX, OY at right angles to one another; on OX take the point A, such that OA = a, and on OY take the point B, such that OB = b. Draw AP, BP perpendicular to OX, OY respectively, meeting in P.

Through P draw any straight line meeting OX, OY in H, K, respectively. Then, if n be the co-ordinates of this line,

η

a relation which is satisfied by the co-ordinates of every line. passing through the point P. This equation therefore represents the point P.

14. In this system, as in that described in the former part of the present chapter, an equation represents the curve, the co-ordinates of whose tangents satisfy it, and an equation of the nth degree will therefore represent a curve of the nth class.

15. If the perpendicular OQ let fall from O on the straight line HK (fig. 23) be denoted by p, and the angle QOX by p, we shall have

and a point will then be represented by the equation

a cos +b sin &=p;

an equation which, if a2 + b2 be put = c2, and becomes p= = ccos (-a).

We thus obtain a method of representing curves by a relation between the perpendicular from a fixed point on the tangent and the inclination of that perpendicular to a fixed straight line. These may be called the tangential polar co