Ex. 4. Given a point and a right line; if OQ be taken inversely as OP, the radius vector to the right line, find the locus of Q. Ex. 5. Given vertex and vertical angle of a triangle and rectangle under sides; if one base angle describe a right line or a circle, find locus described by the other base angle. Take the vertex for pole; let the lengths of the sides be p and p', and the angles they make with the axis 0 and 0', then we have pp' : k2 and 00' = C. = The student must write down the polar equation of the locus which one base angle is said to describe; this will give him a relation between p and 0; then, writing for k2 P, and for 0, C+ ', he will find a relation between p' and ', which will be the polar equation of the locus described by the other base angle. This example might be solved in like manner, if the ratio of the sides, instead of their rectangle, had been given. Ex. 6. Through the intersection of two circles a right line is drawn ; find the locus of the middle point of the portion intercepted between the circles. The equations of the circles will be of the form, p = 2r cos (0 - a); p = 2r' cos (0 - a'); and the equation of the locus will be p = r cos(0 - a) + r' cos (9 — a') ; which also represents a circle. Ex. 7. If through any point O, on the circumference of a circle, any three chords be drawn, and on each, as diameter, a circle be described, these three circles (which, of course, all pass through O) will intersect in three other points, which lie in one right line. (See Cambr. Math. Jour., I. 169.) Take the fixed point O for pole, then if d be the diameter of the original circle; its polar equation will be (Art. 93) d cos 0. In like manner, if the diameter of one of the other circles make an angle a with the fixed axis, its length will be d cos a, and the equation of this circle will be = p = d cosa. cos(0 − a). The equation of another circle will, in like manner, be p = d cos B. cos (0 – (3). To find the polar co-ordinates of the point of intersection of these two, we should seek what value of 0 would render and it is easy to find that 0 must = a + ẞ, and the corresponding value of p = d cos a cosẞ. Similarly, the polar co-ordinates of the intersection of the first and third circles are Now, to find the polar equation of the line joining these two points, take the general equation of a right line, p cos(k − 0) ) = p (Art. 44), and substitute in it successively these values of 0 and p, and we shall get two equations to determine p and k. We shall get p = Hence The symmetry of these values shows that it is the same right line which joins the intersections of the first and second, and of the second and third circles, and, therefore, that the three points are in a right line. * CHAPTER VIII. APPLICATION OF ABRIDGED NOTATION TO THE EQUATION OF THE CIRCLE. 103. If we have an equation of the second degree expressed in the abridged notation explained in Chap. IV., and if we deşire to know whether it represents a circle, we have only to transform to x and y co-ordinates, by substituting for each abbreviation (a) its equivalent (a cosa + y sin a - p); and then to examine whether the coefficient of xy in the transformed equation vanishes, and whether the coefficients of 2 and of y' are equal. This is sufficiently illustrated in the examples which follow. When will the locus of a point be a circle if the product of perpendiculars from it on two opposite sides of a quadrilateral be in a given ratio to the product of perpendiculars from it on the other two sides? Let a = 0, B = 0, y = 0, 80 be the equations of the four sides of the quadrilateral, then the equation of the locus is at once written down ay kßd, which represents a curve of the second degree passing through the angles of the quadrilateral; since it is satisfied by any of the four suppositions, = a = 0, ẞ = 0; a = 0, d = 0; B = 0, y = 0; ẞ = 0, 8 = 0. Now, in order to ascertain whether this equation represents a circle, write it at full length = (x cos a + y sin a-p) (x cos y + y sin y - P.) k(x cosẞ+ y sin ß - p) (x cos d + y sin 8-P). Multiplying out, equating the coefficient of 2 to that of y2, and putting that of xy = 0, we obtain the conditions cos (a + y) = k cos (3 + d); sin (a + y) = k sin (ẞ + d). Squaring these equations, and adding them, we find k = 1; and if this condition be fulfilled, we must have Recollecting that a - ẞ is the angle between the perpendiculars from the origin on the lines a and ß, and is, therefore, the supplement of that angle between a and ß, in which the origin lies, we see that this condition will be fulfilled if the quadrilateral formed by aẞyd be inscribable in a circle (Euclid, 111. 22). And it will be seen on examination that when the origin is within the quadrilateral we are to take k = − 1, and the angle (in which the origin lies) between a and ẞ is supplemental to that between y and 8; but that we are to take k = + 1, when the origin is without the quadrilateral, and the opposite angles are equal. 104. When will the locus of a point be a circle, if the square of its distance from the base of a triangle be in a constant ratio to the product of its distances from the sides? Let the sides of the triangle be a, ß, y, and the equation of the locus is aẞky2. If now we look for the points where the line a meets this locus, by making in it a = 0, we obtain the perfect square y2 = 0. Hence a meets the locus in two coincident points, that is to say (Art. 83), it touches the locus at the point ay. Similarly, ẞ touches the locus at the point ẞy. Hence a and Bare both tangents, and y their chord of contact. Now, to ascertain whether the locus is a circle, writing at full length as in the last article, and applying the tests of Art. 78, we obtain the conditions cos (a + B) = k cos 2y; sin (a + B) = k sin 2y, whence (as in the last article) we get k = 1, a- y = y- ß, or the triangle is isosceles. Hence we may infer that if from any point of a circle perpendiculars be let fall on any two tangents and on their chord of contact, the square of the last will be equal to the rectangle under the other two. Ex. When will the locus of a point be a circle if the sum of the squares of the perpendiculars from it on the sides of any triangle be constant. are The locus is a2 + B2 + y2 = c2: and the conditions that this should represent a circle Squaring and adding, And so, in like manner, each of the other two angles of the triangle are proved to be 60°, or the triangle must be equilateral. 105. To obtain the equation of the circle circumscribing the triangle formed by the lines a = 0, B = 0, y = 0. Any equation of the form β Iẞy + mya + naß = 0, denotes a curve of the second degree circumscribing the given triangle, since it is satisfied by any of the suppositions a = 0, B = 0; B = 0, y = 0; y = 0, a = 0. The conditions that it should represent a circle are found, by the same process as in Art. 103, to be 7 cos (B + y) + m cos (y + a) + n cos (a + B) = 0, Eliminating successively m and n between the equations, we get Now, if C be the angle contained by the sides a, ẞß, then (since aẞ is the angle between the perpendiculars on those sides), hence the equation of the circle circumscribing a triangle is, By sin A+ ya sin B+ aß sin C = 0. 106. The geometrical interpretation of the equation just found deserves attention. If from any point O we let fall perpendicu lars OP, OQ, on the lines a, ß, then (Art. 53) a, ẞ are the lengths of these perpendiculars; and since the angle between them is the supplement of C, the quantity aß sin C is double the area of the triangle OPQ. In like manner, ya sin B and ẞy sin A are double the triangles OPR, OQR. Hence the quan C P tity By sin A+ ya sin B + aß sin C R B is double the area of the triangle PQR, and the equation found in the last article asserts, that if the point O be taken on the circumference of the circumscribing circle, the area PQR will vanish, that is to say (Art. 31, Cor. 2), the three points P, Q, R will lie on one right line. If it were required to find the locus of a point from which, if we let fall perpendiculars on the sides of a triangle, and join their feet, the triangle PQR so formed should have a constant magnitude, the equation of the locus would be and, since this only differs from the equation of the circumscribing circle in the constant part, it is (Art. 78) the equation of a circle concentric with the circumscribing circle. 107. From the equation By sin A+ ya sin B + aß sin C = 0, we can find the equations of the tangents to the circle at the vertices of the triangle. Put the equation into the form y (ẞ sin A + a sin B) + aß sin C = 0, and we saw (in Art. 105) that y meets the circle in the two points where it meets the lines a and B, since, if we make y = 0 in the equation of the circle, that equation will be reduced to aẞ = 0. Now, for the very same reason, the two points in which the line ẞ sin A+ a sin B meets the circle, are the two points where it meets the lines a and B. But these two points coincide, since ẞ sin A+ a sin B passes through the point aß. Hence, since the |