condition (Art. 38), that three lines should meet in a point, may Ex. 1. Find the area of the triangle contained by the three lines la + mß + ny, l'a + &c., &c., (J. J. Walker). Ans. If a, b, c be the sides and ▲ the area of the triangle of reference Ex. 2. The equation of the perpendicular from aʼß'y' on la + mß+ny = 0, may be written 132 (6). The equations of the circle through three points (Art. 94), and of the circle cutting three at right angles (Ex. 2, p. 102), may be written respectively The equation of the latter circle may also be formed by the help of the principle (Ex. 6, p. 102), as the locus of the point whose polars with respect to three given circles meet in a point, in the form x+9′, y+f', g'x +fy + c′ x + g′′, y +f", g′′x +f"y +c"′′ x+g", y+f"" The corresponding equation for any three curves of the second degree will be discussed hereafter. 132 (c). If the radius of a circle vanishes, (x−a)2 + (y—B)2 = 0 the polar of any point x'y', (x′ — a) (x − a) + (y′ − B) (y − B) = 0 evidently passes through the point aß. It is in fact the perpendicular through that point to the line joining aẞ, x'y', as is evident geometrically. Hence then if the circle x2 + y2+2gx+2fy+c=0 reduce to a point, that point which, as being the centre, is given by the equations x+g=0, y+f=0, also satisfies the equation of the polar of the origin gx+fy + c = 0. If given three circles S", S", S"" we examine in what cases IS′+mS"+nS"" can represent a point, we see that the coordinates of such a point must satisfy the three equations 1(x+g')+m(x+g′′)+n (x+g′′′′) = 0, l (y+f') + m (y+f"') + n (y+f'") =0, l(g'x+fy+c)+m(g'′′x +ƒ”y + c'"') + n (g"′′x+ƒ""'y+c") = 0, from which if we eliminate l, m, n, we get the same determinant as in the last article; showing that the orthogonal circle is the locus of all the points that can be represented by IS'+mS"'+nS"". The expression (Ex. 8, p. 103) for the angle at which two circles intersect may be written 2rr' cos 0=2gg +2ff' — c — c'. If now we calculate by the formula of p. 76 the radius of the circle IS'+mS"+nS", and reduce the result by the formula just given, we find (l + m + n)2 r2 = l2x22 + m2r''2 + n2r'''' lg' + mg" + ng"" If'+mf" + nf"" we see + 2mnr'r'"' cos 0′ + 2nlr""'r' cos 0′′ +2lmr′r" cos 0", where ', 0", "" are the angles at which the circles respectively intersect. And since the coordinates of the centre of IS' +mS" +nS"" are that these coordinates will represent a point on the orthogonal circle if l, m, n are connected by the relation r+m2r'”2+&c. =0. If the three given circles be mutually orthogonal this relation reduces itself to its three first terms.* 132 (d). The condition that four circles may have a common orthogonal circle is found by eliminating C, F, G from the four conditions 2Gg+2Ff- C− c = 0, &c., *Casey, Phil. Trans., 1871, p. 586. Since c denotes the square of the tangent from the origin to the first circle, and since the origin may be any point, this condition, geometrically interpreted, expresses (see Art. 94) that the tangents from any point to four circles having a common orthogonal circle are connected by the relation OA.BCD+ OC.ABD = OB". ACD + OD2. ABC.* 132 (e). If a circle x2 + y2+2Gx+2Fy + C=0, cut three others at the same angle 0, we have, besides the equation first given, three others of the form c'+2Rr' cos 0 – 2 Gg′ – 2Fƒ' + C=0; Now if we write 2R cos =λ, the determinant just written is The first determinant equated to zero is, as has just been pointed out, the equation of the orthogonal circle, and the second when expanded will be found to be the equation of the axis of similitude (Art. 117). Thus we have the theorem (Note, p. 109) that all circles cutting three circles at the same angle have a *This theorem is Mr. R. J. Harvey's (Casey, Trans. Royal Irish Acad., XXIV. 458). Since this only differs from the equation of the orthogonal circle by writing c' + λr' for c', &c. we obtain another form for this determinant by making the same change in the last determinant of Art. 132 (6). I owe this form to Mr. Cathcart. common radical axis, viz., the axis of similitude. If in the second determinant we change the sign either of r', r', or r'"', we get the equations of the other three axes of similitude. Now it has been stated (Art. 118) that it is optional which of two supplemental angles we consider to be the angle at which two circles intersect; and if in any line of the first determinant of this article we substitute for its supplement, this is equivalent to changing the sign of the corresponding r. Hence it is evident that we may have four systems of circles cutting the given three at equal angles, each system having a different one of the axes of similitude for radical axis; calculating by the usual formula the radius of the circle whose equation has been written above, we get R in terms of λ, and then from the equation 2R cos 0=λ we get a quadratic to determine the value of corresponding to any value of 0. Ex. 1. To find the condition for the co-existence of the equations ax+by+ c = a'x + b'y + c′ = a′′x + b′′y + c′′ = a""x + b'"'y+c"". Let the common value of these quantities be λ; then eliminating x, y, λ from the four equations of the form ax + by + c = λ, we have the result in the form of a determinant or A+ C = B+ D, where A, B, C, D are the four minors got by erasing in turn each column, and the top row in this determinant. To find the condition that four lines should touch the same circle, is the same as to find the condition for the co-existence of the equations a = ß = y=d. In this case the determinants A, B, C, D geometrically represent the product of each side of the quadrilateral formed by the four lines, by the sines of the two adjacent angles. Ex. 2. The expression, p. 129, for the distance between two points may be written r2 (a sin A + ẞ sinB + y sin C)2 = 0, 0, a β γ Ex. 3. To find the relation connecting the mutual distances of four points on a circle. The investigation is Prof. Cayley's (see Lessons on Higher Algebra, p. 23). Multiply together according to the ordinary rule the determinants which are only different ways of writing the condition of Art. 94; and we get the required relation where (12)2 is the square of the distance between two points. expanded is equivalent to (12) (34) ± (13) (42) ± (14) (23) = 0. This determinant Ex. 4. To find the relation connecting the mutual distances of any four points in a plane. This investigation is also Prof. Cayley's (Lessons on Higher Algebra, p. 24). Prefix a unit and cyphers to each of the determinants in the last example; thus We have then five rows and four columns, the determinant formed from which, according to the rules of multiplication, must vanish identically. But this is (12)2 (34)2 {(12)2 + (34)2 — (13)2 — (14)2 — (23)2 — (24)2} + (13)2 (24)2 {(13)2 + (24)2 — (12)2 — (14)2 — (23)2 — (34)2} +(14)2 (23)2 {(14)2 + (23)2 — (12)2 — (13)2 — (24)2 — (34)2} + (23)2 (34)2 (42)2 + (31)2 (14)2 (43)2 + (12)2 (24)2 (41)2 + (23)2 (31)2 (12)2 = 0. If we write in the above a, b, c for 23, 31, 12; and R+r, R + r', R + r" for 14, 24, 34, we get a quadratic in R, whose roots are the lengths of the radii of the circles touching either all externally or internally three circles, whose radii are r, r', r", and whose centres form a triangle whose sides are a, b, c. Ex. 5. A relation connecting the lengths of the common tangents of any five circles may be obtained precisely as in the last example. Write down the two matrices 1 1, 0, 0, 0, 0 x12 + y'2 — p12, 2x', - 2y', 2r', 1 x''2 + y''2 — p''2, - 2x", - 2y", 2r", 1 &c. 0, 0, 0, 0, &c. |