line (x'y-y'x=0), which joins x'y' to the centre; and its dis any point P is constructed geometrically by joining it to the centre C, taking on the joining line a point M, such that CM.CP=r2, and erecting a perpendicular to CP at M. We see, also, that the equation of the polar is similar in form to that of the tangent, only that in the former case the point x'y' is not supposed to be necessarily on the circle; if, however, x'y' be on the circle, then its polar is the tangent at that point. 89. To find the equation of the polar of x'y' with regard to the ax2+2hxy+by+2gx + 2fy + c = 0. curve We have seen (Art. 86) that the equation of the tangent is ax'x + h(x'y+y'x) + by'y + g(x + x') + f (y + y') + c = 0. This expresses a relation between the coordinates xy of any point on the tangent, and those of the point of contact x'y'. We indicate that the former coordinates are known and the latter unknown, by accentuating the former, and removing the accents from the latter coordinates. But the equation, being symmetrical with respect to the coordinates xy, x'y', is unchanged by this operation. The equation then written above (which when x'y' is a point on the curve, represents the tangent at that point), when x'y' is not on the curve, represents a line on which lie the points of contact of tangents real or imaginary from x'y'. If we substitute x'y' for xy in the equation of the polar we get the same result as if we made the same substitution in the equation of the curve. This result then vanishes when x'y' is on the curve. Hence the polar of a point passes through that point only when the point is on the curve, in which case the polar is the tangent. COR. The polar of the origin is gx+fy+c=0. Ex. 1. Find the polar of (4, 4) with regard to (x−1)2+(y-2)2=13. Ans. 3x+2y=20. Ex. 2. Find the polar of (4, 5) with regard to x2+ y2-3x-4y=8. Ans. 5x+6y=48. Ex. 3. Find the pole of Ax + By + C = 0 with regard to x2 + y2 = p2. Ans. Ar2 (-4, - Br3), as appears from comparing the given equation with xx' + yy' = r2. Ex. 4. Find the pole of 3x + 4y = 7 with regard to x2+ y2 = 14. Ans. (6,8). 90. To find the length of the tangent drawn from any point to the circle (x-a)2 + (y — B)* — y2 = 0. The square of the distance of any point from the centre and since this square exceeds the square of the tangent by the square of the radius, the square of the tangent from any point is found by substituting the coordinates of that point for x and y in the first member of the equation of the circle Since the general equation to rectangular coordinates a (x2+y3) + 2gx + 2fy + c = 0, when divided by a, is (Art. 80) equivalent to one of the form we learn that the square of the tangent to a circle whose equation is given in its most general form is found by dividing by the coefficient of x, and then substituting in the equation the coordinates of the given point. = The square of the tangent from the origin is found by making x and y = 0, and is, therefore, the absolute term in the equation of the circle, divided by a. The same reasoning is applicable if the axes be oblique. #91. To find the ratio in which the line joining two given points x'y', x"y", is cut by a given circle. We proceed precisely as in Art. 42. The coordinates of any point on the line must (Art. 7) be of the form lx" + mx' ly" + my' 1+m > 1+m Substituting these values in the equation of the circle x" + y2 - r2 = 0, and arranging, we have, to determine the ratio 7: m, the quadratic 112 l2 (x'3 + y'' — r2) +2lm (x'x" + y'y" — r3) + m2 (x” +y” — r3) = 0. The values of 1: m being determined from this equation, we have at once the coordinates of the points where the right line meets the circle. The symmetry of the equation makes this method sometimes more convenient than that used (Art. 82). If "y" lie on the polar of x'y', we have x'x" + y'y" — r2 = 0 (Art. 88), and the factors of the preceding equation must be of the form 1 + μm, l— μm; the line joining x'y', x"y" is therefore cut internally and externally in the same ratio, and we deduce the well-known theorem, any line drawn through a point is cut harmonically by the point, the circle, and the polar of the point. *92. To find the equation of the tangents from a given point to a given circle. We have already (Art. 87) found the coordinates of the points of contact; substituting, therefore, these values in the equation xx" +yy" -"=0, we have for the equation of one tangent r (xx' + yy' — x'* − y”) + (xy' — yx') √ (x” + y′′ − r3) = 0, and for that of the other 12 r (xx' + yy' — x'2 — y'2) — (xy' — yx') √(x22 + y′′ — r3) −0. These two equations multiplied together give the equation of the pair of tangents in a form free from radicals. The preceding article enables us, however, to obtain this equation in a still more simple form. For the equation which determines 7: m will have equal roots if the line joining x'y', x"y" touch the given circle; if then "y" be any point on either of the tangents through x'y', its coordinates must satisfy the condition (x2 + y22 − r2) (x2 + y2 — r2) = (xx' + yy' — r2)3. This, therefore, is the equation of the pair of tangents through the point x'y'. It is not difficult to prove that this equation is identical with that obtained by the method first indicated. The process used in this and the preceding article is equally applicable to the general equation. We find in precisely the same way that I m is determined from the quadratic l2 (ax'"* + 2hx"y" + ly" +2gx" +2fy" +c) +2lm {ax'x" + h (x'y" + x'y') + by'y" + g (x' + x′′)+f(y′+y′′)+c} + m2 (ax" + 2hx'y' + by" + 2gx' + 2fy' + c) = 0; from which we infer, as before, that when "y" lies on the polar of a'y' the line joining these points is cut harmonically; and also that the equation of the pair of tangents from x'y' is (ax”2 + 2hx'y' + by'2+2gx' +2fy' + c) (ax2+2hxy+by2+2gx+2fy+c) = {ax'x + h (x'y + xy') + byy' + g (x + x') + ƒ (y + y') + c}2 = 93. To find the equation of a circle passing through three given points. We have only to write down the general equation x2 + y2+2gx+2fy + c = 0, and then substituting in it, successively, the coordinates of each of the given points, we have three equations to determine the three unknown quantities g, f, c. We might also obtain the equation by determining the coordinates of the centre and the radius, as in Ex. 5, p. 4. Ex. 1. Find the circle through (2, 3), (4, 5), (6, 1). Ex. 2. Find the circle through the origin and through (2, 3) and (3, 4). Here c = 0, and we have 13+4g+6ƒ = 0, 25 + 6g+8ƒ=0, whence 2g=-23, 2f=11. Ex. 3. Taking the same axes as in Art. 48, Ex. 1, find the equation of the circle through the origin and through the middle points of sides; and shew that it also passes through the middle point of base. Ans. 2p (x2+ y2) − p (s — s′) x − (p2 + ss′) y = 0. #94. To express the equation of the circle through three points a'y', x"y", x"y" in terms of the coordinates of those points. We have to substitute in x2 + y2+ 2gx + 2fy + c = 0, the values of g, f, e derived from (x2 + y2 )+2gx' + 2fy' +c=0, צווי. (x'''' + y'"'2) + 2gx" +2ƒy"" +c=0. The result of thus eliminating 9, f, c between these four equations will be found to be (x2 + y2 ) {x' (y" — y'") + x" (y"" — y′ ) + x" (y' − y′′ )} - (x2 +y") {x" (y'" − y )+x" ( y − y′′)+x (y′′ - y′′)} +(x"2 + y'2) {x'' (y −y')+x (y' −y")+x′ (y" —y )} - (x'''2+y'''2) {x (y' − y′′) + x′ (y′′ − y )+x" (y − y' )} = 0, as may be seen by multiplying each of the four equations by the quantities which multiply (x + y) &c. in the last written equation, and adding them together, when the quantities multiplying g, f, c will be found to vanish identically. * The reader who is acquainted with the determinant notation will at once see how the equation of the circle may be written in the form of a determinant. 49 If it were required to find the condition that four points should lie on a circle, we have only to write x, y, for x and y in the last equation. It is easy to see that the following is the geometrical interpretation of the resulting condition. If A, B, C, D be any four points on a circle, and O any fifth point taken arbitrarily, and if we denote by BCD the area of the triangle BCD, &c., then ОA2. BCD + OC2.ABD=0B“.ACD +OD2. ABC. 95. We shall conclude this chapter by showing how to find the polar equation of a circle. We may either obtain it by substituting for x, p cose, and for y, p sin (Art. 12), in either of the equations of the circle already given, a (x2 + y2) + 2gx + 2fy+c=0, or (x − a)2 + (y − ß)2 = r2, or else we may find it independently, from the definition of the circle, as follows: Let O be the pole, C the centre of the circle, and OC the fixed axis; let the distance OC=d, and let OP be any radius vector, and, therefore, = p, and the angle POC=0, then we have that is, or P' P C PC2=OP2+OC-20P.OC cos POC, r2 = p2 + d2 - 2pd cose, p*- 2dp cos 0+ď3 — r2 = 0. This, therefore, is the polar equation of the circle. If the fixed axis did not coincide with OC, but made with it any angle a, the equation would be, as in Art. 44, p2 - 2dp cos (0 - a) + ď2 — r2 = 0. If we suppose the pole on the circle, the equation will take a simpler form, for then rd, and the equation will be reduced to p = 2r cos 0, a result which we might have also obtained at once geometrically from the property that the angle in a semicircle is right; or else by substituting for x and y their polar values in the equation (Art. 79) x* + y2 = 2rx. |