(3) We would also advise the student to obtain each of these equations independently from their appropriate figures. Work like this will supply him with the most valuable kind of exercises, and test his knowledge most thoroughly. (4) The general form of the equation leaves the origin of the axes quite unrestricted as to its position so long as it is in the plane of the circle. The student may, therefore, wisely employ himself in obtaining the equation after placing the origin on different sides of the circle and in different parts within the circle. (5) The last three varieties of the equation in (2) imply that the origin is on the curve. In these cases there can be no absolute term, that is, one which contains neither x nor y; for the equation must be satisfied by the values x = 0, y=0. (6) The student should inquire what the results will be in supposing the circle to fulfil different conditions. When, for instance, (a) the circle cuts one or both of the axes at given points, (b) touches one or both of the axes at a given point or points. We will give a few examples of this kind of problems. Ex. 1. A circle passes through the origin and intercepts lengths h and k respectively from the positive parts of the axes of x and y. Find the equation to this circle. Ex. 2. Find the equation to a circle referred to two tangents at right angles to each other as axes. Ex. 3. A circle in a plane has its centre at the point whose rectangular co-ordinates are a and b, and passes through the origin. Find its equation, that of the tangent to it at the origin, and the lengths of the intercepts it cuts off on the axes. Ex. 4. On the straight line joining (x, y1) and (x2 y2) as a diameter a circle is described. Find its equation. These are the best class of exercises for the Pass candidate. Ex. 3 was given in the year 1875. (7) The simplest form of equation is x2+y2=c2; and this is the form in which the equation to the circle is usually expressed when we can choose rectangular axes and are at liberty to place the origin of the axes at any point we please. It is also the form most generally used; and the student should observe that any results obtained for this form can be put in the general form by writing x-a for x, and y-b for y. (8) By writing the equation x2+ y2 − 2cx=0 in the proportional form xyy: 2c-x, we may infer the geometrical property of the circle that a perpendicular from any point in the circumference of a circle to a diameter is a mean proportional between the two parts into which the perpendicular divides the diameter. (9) By writing it in the form x2+y2=2cx, and observing that x2+y2= the square of the distance between the origin and any point in the circumference, we may infer that a chord drawn from one extremity of a diameter is a mean proportional between the diameter itself and that remote part of it which the perpendicular from the extremity of the chord cuts off. 38. To find the equation to the circle referred to oblique axes. The student will seldom have need to make use of this equation. It reduces to the form for rectangular axes when the cosine of the angle of the inclination of the axes is equal to zero, that is, when cosw = 0. When the oblique axes have their origin at the centre of the circle the equation has the form x2 + y2+2xy cosw=c2. Here also we get x2 + y2=c2 when cosw=0, that is, when w=90°. The General Form of Equation. 39. Show that the general equation to the circle, for rectangular axes, is always of the form x2 + y2+Dx + Ey+F=0, where D, E, F are constant quantities, any one or more of which in particular cases may be equal to zero. This may be shown by examining the different forms which the equation may assume, as given in Article 37, (1) and (2), and comparing them with the general form which evidently includes them all. Observe particularly, that whenever we obtain an equation of the second degree which, after reduction, has unity as the co-efficient of each of the squares of x and y, the variables, and does not contain the product of x and y, we may, in general, assume it to be the equation to a circle. It is well, however, to examine the equation; for it may in some instances represent the locus of a point, or of an imaginary curve, as shown in the next article. If the equation is of the form Ax2+Ay2+Dx + Ey+F=0, it may E F Ꭰ х be reduced to the form x2+y2+x+y+=0, by dividing by A. This is the most general form which the equation can assume with rectangular axes. But the form given in the proposition is the one usually employed. 40. Show that the locus of the general equation x2+y2+Dx + Ey + F=0 is always the circumference of a circle, including, however, the peculiarities of a circumference reduced to a mere point and of a curve merely imaginary. Completing the squares of x2+Dx and y2+ Ey, we may write the above equation thus: E2 D 2 x2 + Dx + (22)2 + y2+ Ey + (1)2 = ( 2 ) + (1)2 - 1 or, (x+D)2+(y+}E)2 = {(D2 + E2-4F). = F; (a) By comparing this with the form (x − a)2+(y-b)2=c2, we shall see that it represents a circle whose radius √(D2+E2 - 4F), and the co-ordinates of whose centre are -D and E. In this case D2+ E2-4F must be positive. (b) If D2+ E2 - 4F be negative, there are no real values of x and y which can satisfy the equation, and the circle is imaginary. In other words, if 4F be greater than D2+ E2 the locus is said to be impossible; for then (D2+ E2-4F) will be the square root of a negative quantity, which is, of course, imaginary. (c) If D2+ E2-4F=0, the sum of the squares on the left-hand side of the equation will be zero, and we have the simultaneous values xD and y= - E, which are the co-ordinates of a point. This point may be regarded as the circumference of a circle whose radius is indefinitely small. It may also be considered as the point of intersection of the two imaginary lines x+D+√ −1(y+†E)=0, x+†D−√√ − 1 (y + }E) = 0. We may suppose these two lines to intersect in the only real point for which the equation in case (c) will be satisfied. But this is an aspect of the case to which the reader need not give much attention except as showing the consistency of Analytical Geometry throughout. The student will be often asked to construct the circles represented by certain equations; that is, to find the radius and the co-ordinates of the centre. This process in its general applications is sometimes called determining the loci of the given equations. Ex. 1. Construct the locus of the equation 2x2 + 2y2 - 3x+4y-1=0. We assume that the locus is a circle, for the product of the variables does not occur, and the squares of the variables have equal coefficients. Dividing the equation by 2, the equation becomes x2+ y2 - 3x+2y=0, completing the squares with x2-x and y2+2y, we have x2-x+(2)2+ y2+2y+1=1+1+}={}; or, (x)+(y+1)2=8. Comparing this with the form (x − a)2 + (y − b)2=c2, we find that the radius of the circle is √33, and the co-ordinates of its centre and -1. Ex. 2. Interpret the equation x2 + y2+2x+6y+11=0.. We may write it (x+1)2+(y+3)2= -1; whence we see that the equation has no geometrical meaning, or that the locus is impossible. Here the radius is the imaginary quantity √-1, and therefore the circle is imaginary, or has no real existence. Ex. 3. Find the locus of the equation 4x2+4y2 - 12x - 8y +13=0. Dividing by 4 and completing the squares we have, after reduction, (x − 3)2 + (y − 1)2 = 0. Hence the given equation represents a point whose co-ordinates are (3, 1); or, in other words, a circle whose radius = 0, and the co-ordinates of whose centre are x=3, y=1. Sometimes equations to the circle are proposed far more complicated than the above; but they may always be reduced to the form, x2 + y2 + Dx + Ey+F=0, when they are referred to rectangular axes. The more general form, which includes both kinds of axes, is given in the next article. 41. To find the condition that the general equation of the second degree, Ax2+ Bxy + Cy2 + Dx + Ey+F=0, may represent a circle when referred (1) to rectangular axes, (2) to oblique axes. Suppose, for the time being, that this is the equation to the circle whose radius = c and whose centre is the point (a b). (1) When the axes are rectangular, the equation must be identical with x2+y2 - 2ax − 2by + a2 + b2 − c2 = 0. Dividing by A, and equating the coefficients of xy, y2, x, y, and the absolute terms, we have B=0,=1, =a2+b2-c2. D = 2a, E = F A 26, Α Putting these values in the equation of (1), and observing that the value of B is zero, we shall have the equation Ax2+Ay2+Dx + Ey+F=0, which, when divided by A, corresponds to the general equation of Articles 39 and 40, and therefore the condition that the general equation of the second degree shall represent a circle is that B=0, and C=A. (2) When the axes are oblique, the proposed general equation must be identical with x2 + y2+2cosw xy − 2 (a+b cosw)x− 2(b+a cosw)y+a3 +b2+ 2abcosw-c2=0, which is what the equation referred to oblique co-ordinates becomes when expanded and rearranged. By a process similar to that in (1) we shall find that the general equation to the circle in this case is Ax2+Ay2+2A cosw xy + Dx + Ey+F=0. Hence, in order that the general equation of the second degree may represent a circle, the condition is that B=2A cosw and C=A, where is the inclination of the axes. If the inclination of the axes be 90°, that is, if w= π then cosw =0 and B=0, which makes the result in (2) correspond with (1), as it evidently ought to do. THE CIRCLE UNDER PARTICULAR CONDITIONS. 42. To find the conditions that two circles may (1) cut each other or (2) touch each other. We will briefly indicate the method of proof. Let the equations to the circles be x2+y2 = c2, and (x − d)2+ y2 = c12, where d is the distance between their centres. Treating these as simultaneous equations, we shall find that x=. (c2 - c12 + d2), 2a√ {4d2 c2 - (c2 − c 12 + d2) 3 } ; and y = ± or y = ± 1 1 2 d 2d √{(c+d+c1} (c+d− c1) (c1+c−d) (c1−c+d)}. (1) If in the last result there be no negative factor, the expression for y will be real, and the two values of y will be equal, but of opposite signs. Hence the circles will intersect in two points. (2) If any of the factors be equal to zero, then y=0, and the circles will touch one another, for they will have one point, and only one point, in common. This point is given us by the value of x, and will lie on the line joining the centres. 43. To find the equation to the straight line which passes through the intersection of two circles which cut each other. Let the equations to the circles be x2+y2+Dx+Ey+F=0...(1), and x2+y2+D1x+E1y+F1 =0...(2) Subtract (2) from (1), then (D-D1)x+(E-E1)y+F-F1 =0...(3) 1 This is the equation to the straight line required. For it is an equation of the first degree between the two variables x and y; and it passes through the intersection of (1) and (2), for whenever the equations (1) and (3) are satisfied simultaneously, the equation (2) is also satisfied. 44. To find the equation to the circle which passes through the points of intersection of two circles which cut each other. Let the equations to the two circles be as before. Then, if k be any constant, x2 + y2+Dx + Ey +F+k(x2 + y2+ D1x + E1y + F1)=0. (1) is the equation to some locus which passes through the points of intersection of the two circles; for the equation is satisfied by those values of x and y, which also satisfy the equations to the two circles. The above equation may be put in the form = 0... (2) (1 + k) x2 + (1 + k) y2 + (D+kD1)x+(E+kE1)y+F+kF1 which is obviously the equation to a circle. (See Article 39.) NOTE. Suppose we wish to find the equation to a circle which not only passes through the above points of intersection, but also through the particular point (x1y1). Since the circle passes through this point, the values x=x1, y=y1 will satisfy equation (1). Substitute these values in (1); we shall then have an equation which S |