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:. b2 (q− p)2 + c2 (r− p)2 — (b2+ c2—a2) (q − p) (r− p) = b2c2 sin3A,

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or a2(p−q) (p − r) + b2 (q− r) (q − p) + c2 (r − p) (r — q) = 4A2.

This is the relation between the co-ordinates of any line whatever.

COR. Since the line at infinity may be considered as equidistant from A, B, and C, it will be represented by the equations p=q=r.

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5. An equation of a degree, higher than the first, may be regarded as representing the curve which is touched by all the straight lines, the co-ordinates of which satisfy the equation of the curve. Adopting this mode of interpretation, the values of the ratios p q r which simultaneously satisfy two given equations will be the co-ordinates of the common tangents to the two curves represented by these equations, and the values obtained by combining any given equation with an equation of the first degree, will represent all the straight lines which pass through the point represented by the equation of the first degree, and which touch the curve. From this it follows, that an equation of the nth degree will represent a curve such that n tangents, real or imaginary, can be drawn to it from any point, that is, a curve of the nth class.

It will hence follow that every equation of the second degree represents a conic. We may proceed to consider some of its more interesting special forms.

6. To find the equation of a conic which touches the three sides of the triangle of reference.

are

The co-ordinates of the sides of the triangle of reference

q=0, r=0 for BC,

r=0, p=0 for CA,

p=0, q=0 for AB.

Hence, the equation of the required conic must be satisfied whenever two out of the three co-ordinates p, q, r are = 0. It must therefore be of the form

Lqr +Mrp+Npq=0.

The equations of the points of contact are

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These may be established as follows: If in the given equation we make Mr+ Nq=0, we obtain either q = 0, or r=0. It hence appears that the tangents drawn through the point Mr+Nq=0, pass either through the point q=0, or through the point r = 0. 0. But the three points

Mr+ Nq=0, q=0, r=0,

lie in the same straight line; hence the tangents drawn from Mr+Nq=0 coincide, that is, it is the point of contact of the tangent for which q=r=0. Similarly for the other two points of contact.

It will hence appear, by reference to the equations of the points of contact of the inscribed circle, given in Art. 2, that that circle is represented by the equations

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7. To find the equation of a conic circumscribed about the triangle of reference.

The equations of the angular points of the triangle of reference are p=0, q=0, r=0. Now, since each of these points lies on the curve, the two tangents drawn through any one of them must coincide, hence when any one of these quantities is put = 0, the remaining equation must have two equal roots. The required equation will therefore be of the form

L2p2 + M3q' +N22 - 2MNqr-2NLrp-2LMpq=0.

The co-ordinates of the several tangents at the angular points will be given by the equations

p=0, Mq-Nr=0,

q=0, Nr - Lp = 0,

r=0, Lp-Mq=0.

If the conic be a circle, the tangent at A will be determined by the equations

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which last is equivalent to b2q — c2r = 0.

Similar equations holding for the other two tangents, the equation of the circumscribing circle will be

a*p2 + b1q2 + c3r2 — 2b3c3qr — 2c3a3rp — 2a2b3pq = 0,

which may be reduced to

± ap1±bq1 ± cr3 = 0.

8. By investigations similar to those in Chap. IV. Art. 8, it may be shewn that the equation of the pole of the line (f, g, h) with respect to the conic

is

$ (p, q, r) = up2 +vqa + wr2 + 2u'qr +2v'rp + 2w'pq = 0,

(uf+w'g+v'h) p + (w'f+vg+u'h) q + (v'f+u'g+wh) r = 0. Now, the centre is the pole of the line at infinity, which is given by the equations p=q=r.

The equation of the centre is therefore

(u + v' + w') p + (u' + v + w') q + (u' + v' + w) r = 0.

If the conic be a parabola, it touches the line at infinity; the condition that it should be a parabola is therefore

u+v+ w + Qu' + 2v′ + 2w' = 0.

9. The two points in which the conic is cut by the line (f, g, h) are represented by the equation

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Hence, the two points in which it is cut by the line at infinity are given by

4(u+v+w+2u'+2v'+2w') (up2+vq2+wr2+2u'qr+2v'rp+2w'pq) −{(u+v'+w') p+(u'+v+w') q+(u'+v'+w)r}2=0.

Hence may be deduced the equation of the two points at infinity through which all circles pass. For these are the same for all circles. Now, for the inscribed circle they are obtained by putting

= 8-b,

2w's - c.

u = v=w= 0,
The equation then becomes

Qu' = s― α, 2v' =

4s {(s—a) qr + (s− b) rp + (s — c) pq} − (ap+bq+cr)2=0,

or a2p2+b2q2+c2r2—2bcgr cos A-2 carp cos B-2abpq cos C=0,

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.

10. Every circle may therefore be represented by the equation

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

− (lp +mq + nr)2 = 0,

where lp+mq+nr=0 is the equation of the centre.

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, we get

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

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a2 — uk = l2, b2 - vk = m2, c2 - wk = n2,

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(b2 — vk) (c2 — wk) = (bc cos A + u'k)o,

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

Again, we get

(a2 —uk) (bc cos A + u'k) + (ca cos B+ v'k) (ab cos C+w'k) =0, or (v'w'-uu') k2+{a (au'+bcos C.v'+ccos B.w')- becos A.u} k -a2bc sin B sin C=0.

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