The reciprocity of tangential and ordinary equations will be better seen if we solve the converse problem, viz. to find the equation of the curve, the tangents to which fulfil the condition. X”a+μ”ß+v′′y be any two lines, such that X''v, X"μ"v" satisfy the above condition, and which therefore are tangents to the curve whose equation we are seeking; then is the tangential equation of their point of intersection. For (Art. 70) any equation of the form A+ Bu+ Cv=0 is the condition that the line λa+uß+vy should pass through a certain point, or, in other words, is the tangential equation of a point; and the equation we have written being satisfied by the tangential coordinates of the two lines is the equation of their point of intersection. Making X', μ', v′ = X", μ", v" we learn that if there be two consecutive tangents to the curve, the equation of their point of intersection, or, in other words, of their point of contact, is Solving for X', u', v' from these equations, and substituting in the relation, which by hypothesis X'u'v satisfy, we get the required equation of the curve √(la) +√(mß) + √√(ny) = 0. 131. The conditions that the equation of Art. 129 should represent a circle are (Art. 128) m2 sin C+ n' sin'B+2mn sin B sin C=n" sin A+ sin" C +2nl sin C sin A = l2 sin3 B+ m2 sin*A +2lm sin A sin B, or m sin C+n sin B=+ (n sin A + 1 sin C) = + (l sin B+m sin A). Four circles then may be described to touch the sides of the given triangle, since, by varying the sign, these equations may be written in four different ways. If we choose in both cases the sign, the equations are I sin C-m sin C+ n (sin A - sin B) = 0; 7 sin B+m (sin A – sin C) – n sin B = 0. The solution of which gives (see Art. 124) l=sin A (sin B+ sin C- sin A), m=sin B (sin C+ sin A-sin B), n = sin C (sin A+ sin B- sin C). But since in a plane triangle sin B+ sin C - sin A4 cos A sin B sin C, these values for l, m, n are respectively proportional to cos3A, cos B, cos2C, and the equation of the corresponding circle, which is the inscribed circle, is cos A√(a) + cos †B √(B) + cos 1⁄2 C √(y) = 0,* or a2 cos11A+82 cos*}B+ y2 cos*†C-2By cos*B cos - 2ya cos2C cos31⁄2A – 2aß cos31⁄2 A cos2+B=0. We may verify that this equation represents a circle by writing it in the form * Dr. Hart derives this equation from that of the circumscribing circle as follows: Let the equations of the sides of the triangle formed by joining the points of contact of the inscribed circle be a′ = 0, ẞ′ = 0, y′ = 0, and let its angles be A', B', C'; then (Art. 124) the equation of the circle is B'y' sin A' + y'a' sin B' + a'ß' sin C' = 0. But (Art. 123) for every point of the circle we have a" = ẞy, ß" = ya, y'2 = aß, and it is easy to see that A' = 90 — §A, &c. Substituting these values, the equation of the circle becomes, as before, cos A √(a) + cos &B √(B) + cos &C √(y) = 0. If the equation of the note, p. 119, be treated similarly, we find that every point of the circle, of which a, ß, y, ô are tangents, satisfies the equation, cos (12) cos (23) cos (34) cos (41) + + + (aß) J(BY) √(28) √(da) = 0, where (12) denotes the angle between aß, &c. Similarly for any number of tangents. In the same way, the equation of one of the exscribed circles is found to be a2 cos⭑A+ B2 sin*B + y2 sin* C-2By sin2 B sin' C or +2ya sin2C cos2 4+2aß sinВ cos*}A = 0, cos √(− a) + sin }B √√(ß) + sin } C √(y) = 0. The negative sign given to a is in accordance with the fact, that this circle and the inscribed circle lie on opposite sides of the line a. Ex. Find the radical axis of the inscribed circle and the circle through the middle points of sides. The equation formed by the method of Art. 128 is 2 cos2 4 cos2B cos2C (a cos A+ ẞ cos B + y cos C'} Divide by 2 cos A cos B cos or C, and the coefficient of a in this equation is cos 4 (2 cos2 4 sin B sin C-cos A cos B cos &C'}, The equation of the radical axis then may be written and it appears from the condition of Art. 130 that this line touches the inscribed circle, the coordinates of the point of contact being sin2 (B-C), sin2 (C-A), sin2 (A— B). These values shew (Art. 66) that the point of contact lies on the line joining the two centres whose coordinates are 1, 1, 1, and cos (B - C'), cos (C-A), cos (A – B). In the same way it can be proved that the circle through the middle points of sides touches all the circles which touch the sides. This theorem is due to Feuerbach.* * Dr. Casey has given a proof of Feuerbach's theorem, which will equally prove Dr. Hart's extension of it, viz. that the circles which touch three given circles can be distributed into sets of four, all touched by the same circle. The signs in the following correspond to a triangle whose sides are in order of magnitude a, b, c. The exscribed circles are numbered 1, 2, 3, and the inscribed 4; the lengths of the direct and transverse common tangents to the first two circles are written (12), (12)'. Then because the side a is touched by the circle 1 on one side, and by the other three circles on the other, we have (see p. 115) showing that the four circles are also touched by a circle, having the circle 4 on one side and the other three on the other. 2 x 3 1/22 132. If the equation of a circle in trilinear coordinates is equivalent to an equation in rectangular coordinates, in which the coefficient of x2 + y2 is m, then the result of substituting in the equation the coordinates of any point is m times the square of the tangent from that point. This constant m is easily determined in practice if there be any point, the square of the tangent from which is known by geometrical considerations; and then the length of the tangent from any other point may be inferred. Also, if we have determined this constant m for two circles, and if we subtract, one from the other, the equations divided respectively by m and m', the difference which must represent the radical axis will always be divisible by a sin A+B sin B+ y sin C. Ex. 1. Find the value of the constant m for the circle through the middle points of the sides a2 sin A cos A+ ß2 sin B cos B + y2 sin C cosC – ẞy sin A — ya sin B – aß sin C = 0. Since the circle cuts any side y at points whose distances from the vertex A are te and b cos A, the square of the tangent from A is be cos A. But since for A we have B=0, y = 0, the result of substituting in the equation the coordinates of A is a" sin A cos A (where a' is the perpendicular from A on the opposite side), or is bc sin A sin B sinC cos A. It follows that the constant m is 2 sin A sin B sin C. Ex. 2. Find the constant m for the circle ẞy sin A + ya sin B + aß sinC. If from the preceding equation we subtract the linear terms (a cos A + ẞ cos B+ y cos C) (a sin A + ẞ sin B + y sinC), the coefficient of x2+ y2 is unaltered. The constant therefore for By sin A &c., is – sin A sin B sin C. It follows that for an equation written in the form at the end of Art. 128 the constant is - k sin A sin B sin C. Ex. 3. To find the distance between the centres of the inscribed and circumscribing circle. We find D2 - R2, the square of the tangent from the centre of the inscribed to r2 (sinA+sin B+sin C) the circumscribing circle, by substituting a=ẞ=y=r, to be = or, by a well-known formula, - 2Rr. Hence D2 R2 - 2Rr. sin A sin B sin C Ex. 4. Find the distance between the centres of the inscribed circle and tha, 'hrough the middle points of sides. If the radius of the latter be p, making use of the formula, we have sin cos A + sin B cos B + sin C cos C = 2 sin A sin B sin C, Assuming then that we otherwise know R = 2p, we have Dr-p; or the circles touch. Ex. 5. Find the constant m for the equation of the inscribed circle given above. Ans. 4 cos24 cos2 B cos2 C Ex. 6. Find the tangential equation of a circle whose centre is a'ß'y' and radius r. This is investigated as in Art. 86, Ex. 4; attending to the formula of Art. 61; and is found to be (\a' + μß' + vy')2 = p2 (A2 + μ2 + v2 – 2μv cos A – 2vλ cos B - 2λμ cos C'). The corresponding equation in a, ß, y deduced from this by the method afterwards explained, Art. 285, and is r2 (a sin A+ ẞ sin B + y sin C)2 = (By' — ẞ'y)2 + (ya' — y'a)2 + (aß′ – a′ß)2 −2 (ya'—y'a) (aß′ — a′ß) cos.A — 2 (aß'—a'ß) (By'—ẞ'y) cosB-2 (By' —ẞ'y) (ya'—y'a) cosC. This equation also gives an expression for the distance between any two points. Ex. 7. The feet of the perpendiculars on the sides of the triangle of reference from 1 1 the points a', B', '; '''; (see Art. 55) lie on the same circle. By the help of Ex. 6, p. 60, its equation is found to be (By sina+ya sin B+aß sin C) (a' sinA+ß′ sinB+y' sin C)(B′y'sinA+y'a' sinB+a'ß' sinC) =sin A sin B sin C (a sin A + B sin B + y sin C) (aa'(B′+y'cosA)(y′+ẞ'cos▲) ßß'(y'+a' cosB) (a'+y'cosB) ̧yy′(a'+ß'cosC)(B'+a'cos C')) sin A + Ex. 8. It will appear afterwards that the centre of a circle is the pole of the line at infinity a sin A + ẞ sin B + y sin C; and it is evident that if we substitute the coordinates of the centre in the equation of a circle, for which the coefficient of x2+ y2 has been made unity, we get the negative square of the radius. By these principles we establish the following expressions of Mr. Cathcart. The coordinates of the centre of the circle (Art. 128) R are (la + mẞ + ny) (a sin A+ &c.) + k (By sin A + &c.), R (k cos A + 1− m cosC - n cos B), (k cos B-l cosC + m − n cos A), where R is the radius of the circumscribing circle. The radius p is given by the equation k2p2 = R2 {k2 + 2k (l cos A + m cos B + n cosC) + 12 + m2 + n2 - 2mn cos A – 2nl cos B-2lm cos C}, and the angle of intersection of two circles is given by =1+ pp' cos 0 R2 + Icos A+ m cos B+ n cos C l'cos A+ m' cos B + n' cosC + k k' ll' + mm' + nn' — (mn' + m'n) cos A − (nľ′ + n'l) cos B — (Im' + I'm) cosC 132(a). In the earlier editions of this book I did not venture to introduce the determinant notation, and in the preceding pages I have not supposed the reader to be acquainted with it. But the knowledge of determinants has become so much more common now than it was, that there seems no reason for excluding the notation, at least from the less elementary chapters of the book. Thus the equation of the line joining two points (Art. 29), the double area of a triangle (Art. 36) and the S |