CHAPTER V. = 0, and (2) = 0, - r. + y z + I>3c +Ey + F= O. D = - 2 a, E = - 2 ft, THE CIRCLE AND THE EQUATION. -E 2 to both

Size: px

Start display at page:

Download "CHAPTER V. = 0, and (2) = 0, - r. + y z + I>3c +Ey + F= O. D = - 2 a, E = - 2 ft, THE CIRCLE AND THE EQUATION. -E 2 to both"

Jewel Sullivan
5 years ago
Views:

1 CHAPTER V THE CIRCLE AND THE EQUATION 2 + y z + I>3c +Ey + F= O 52. The general equation of the circle. If (a, ft) is the center of a circle whose radius is r, then the equation of the circle is (Theorem II, p. 51) (1) x 2 + y 2-2 ax - 2/3y + a? + ft 2 - r 2 = 0, or (2) (v-af + d, -?) =,*. In particular, if the center is the origin, a = 0, ft = 0, and (2) reduces to (3) x z + y* = r\ Equation (1) is of the form (4) x 2 + if + Dx + Ey + F = 0, where (5) D = - 2 a, E = - 2 ft, and F = a 2 + f? - r 2. Can we infer, conversely, that the locus of every equation of. the form (4) is a circle? By adding J D 2 + \ -E 2 to both members, (4) becomes (6) (x + %Dy+(y+iEy = i(d 2 + E 2 -F). In (6) we distinguish three cases: If D 2 -f- E' 2 4 F is positive, (6) is in the form (2), and hence the locus of (4) is a circle whose center is ( J Z), J E) and whose radius is r = Vz> 2 + E 2 4 F. If D 2 + E 2 &F = 0, the only real values satisfying (6) are x = ^D, y \E (footnote, p. 52). The locus, therefore, is the single point ( Z>, J E). In this case the locus of (4) is often called a point-circle, or a circle whose radius is zero. If D 2 + E 2 4 F is negative, no real values.satisfy (6), and hence (4) has no locus. 115

2 116 ANALYTIC GEOMETRY The expression D 2 + E 2 4 F is called the discriminant of (4), and is denoted by. The result is given by Theorem I. The locus of the equation (I) v? + y 2 + Vx + Ey + F = 0, whose discriminant is = D 2 + E z ~ 4 F, is determined as follows : (a) When is positive the locus is the circle whose center is (%D, %E) and whose radius is r = Vl>2 + E' 2 4 F = % V. (b) When 1*5 zero the locus is the point-circle ( Z), ^ ). (c) When ts negative there is no locus. Corollary. When E = /i<? center of (I) is on ^e X-axis, and when D = /ie center is on the Y-axis. Whenever in what follows it is said that (I) is the equation of a circle it is assumed that is positive. Ex. 1. Find the locus of the equation x 2 + y 2 4x + 8y 5 = 0. Solution. The given equation is of the form (I), where f*, \ \ A' D = - 4, E = 8, F = - 5, and hence The locus is therefore a circle whose center is the point (2, 4) and whose radius is \ VlOO = 5. The equation Ax 2 + Bxy + Cy 2 -f Dx -f Ey + F = is called the general equation of the second degree in x and y because it contains all possible terms in x and y of the second and lower degrees. This equation can be reduced to the form when (I) and only when A = C and 5 = 0. Hence the locus of an equation of the second degree is a circle only when the coefficients of x 2 and y 2 are equal and the xy-term is lacking. 53. Circles determined by three conditions. The equation of any circle may be written in either one of the forms or

3 THE CIRCLE 117 Each of these equations contains three arbitrary constants. To determine these constants three equations are necessary, and as any equation between the constants means that the circle satisfies some geometrical condition, it determined to satisfy three conditions. follows that a circle may be Rule to determine the equation of a circle satisfying three conditions. First step. Let the required equation be or as m,ay be more convenient. Second step. Find three equations between the constants a, ft and r [or D, E, and F~\ which express that the circle (1) [or (2)] satisfies the three given conditions. Third step. Solve the equations found in the second step for a, ft and r \_or D, E, and F]. Fourth step. Substitute the results of the third step in (1) [or (2)]. The result is the required equation. Ex. 1. Find the equation of the circle passing through the three points Pi(0, 1), P 2 (0, 6), and P 8 (3, 0). Solution. First step. Let the required equation be (3) x 2 + y 2 + Dx + Ey + F = 0. Second step. Since PI, P 2, and P 3 lie on (3), their coordinates must satisfy (3). have Hence we -3 N (4) 1 +.E + ** =.(), (5) E + F = 0, and (6) ^ = 0. Third step. Solving (4), (5), and (6), we obtain E = - 7, F = 6, D = - 5. Fourth step. Substituting in (3), the required equation is

4 118 ANALYTIC GEOMETRY By Theorem I we find that the radius is point (, ). f V2* and the center is the Ex. 2. Find the equation of the circle passing through the points PI (0, 3) and P 2 (4, 0) which has its center on the line x + 2 y = 0. Solution. First step. Let the required equation be (7) x 2-1- y 2 + Dx + Ey + F = 0. Second step. Since PI and P 2 lie on the locus of (7), we have (8) 9-3^ + ^ = and (9) 16

5 THE CIRCLE Find the locus of the following equations. (a) x2 + 2/2 _ Gx _ 16 = 0. (f) x2 + 2/2 _ Q X + y _ 5 _ o. (b) 3x2 + 3y2-10 x - 24 y = 0. (g) (x + I) 2 + (y - 2) 2 = 0. (c) x2 + y 2 = 0. (h) 7x y 2-4x - y = 3. (d) x2 + y 2-8x - Qy + 25 = 0. (i) x 2 + y ox + 2 fy + a 2 + ft2 = 0. (e) x2 + y 2-2x + 2 y + 5 = 0. (j) x 2 + y' + IGx = Find the equation of the circle which (a) has the center (2, 3) and passes through (3, - 2). Ana. x 2 + y 2-4 x - 6 y - 13 = 0. ;-$>) passes through the points (0, 0), (8, 0), (0, - 6)., Ana. x 2 + / 2 _8x + 6y = 0. (c) passes through the points (4, 0), ( 2, 5), (0, - 3). Ans. 19x / 2 + 2x- 47 y- 312 = 0. (d) passes through the points (3, 5) and (3, 7) and has its center on the JT-axis. Ans. x 2 + y x - 46 = 0. (e) passes through the points (4, 2) and (6, 2) and has its center on the T-axis. Ans, x 2 + 2/2 + 5 y - 30 = 0. (f) passes through the points (5, 3) and (0, 6) and has its center on the line 2x-3y-6 = 0. Ans. 3 x y x - 64 y = 0. (g) has the center (1, 5) and is tangent to the X-axis. Ans. x 2 + 2/2 _ _ 2 x + 10 y + 1 = 0. (h) passes through (1, 0) and (5, 0) and is tangent to the F-axis. Ana. x 2 +?/ 2-6x;t2V5y + 5 = 0. (i) passes through (0, 1), (5, 1), (2, - 3). Ans. 2x 2 + 2?/ 2-10x + y-3 = 0. has the (j) line joining (3, 2) and (7, 4) as a diameter. Ans. x 2 + y x - 6 y 13 = 0. (k) has the line joining (3, - 4) and (2, 5) as a diameter. Ans. x 2 + y 2 _ 5 x + 9 y + 26 = 0. (1) which circumscribes the triangle formed by x 6 = 0, x + 2?/ = 0, and x - 2 y = 8. Ans. 2 x y 2 _ 21 x + 8 y + 60 = 0. (m) passes through the points (1, 2), (- 2, 4), (3, - 6). Interpret the result by the Corollary, p. 89. (n) is inscribed in the triangle formed by4x + 3y 12 = 0, y 2 = 0, -10 = 0. Ana. 36x y 2-516x + QQy = Plot the locus of x 2 + y 2-2x + 4 y + k = for k = 0, 2, 4, 5-2, - 4, 8. What values of k must be excluded? Ans. k > 5.

6 120 ANALYTIC GEOMETRY 5. What is the locus of x 2 + y 2 + Dx + Ey + F if D and E are fixed and F varies? 6. For what values of k does the equation x 2 + y 2 4 jc + 2 ky + 10 = have a locus? Ans. k > -f Vo and A: < - V6. 7. For what values of fc does the equation x 2 + y 2 + kx + F=Q have a locus when (a) F is positive ; (b) F is zero ; (cj F is negative? ^Ins. (a) fc > 2 \/F and fc < - 2 V* 1 ; (b) and (c) all values of k. 8. Find the number of point-circles represented by the equation in problem 7. Ans. (a) two ; (b) one ; (c) none. 9. Find the equation of the circle in oblique coordinates if w is the angle between the axes of coordinates. Ans. (x - 2 a) + (y - 2 /3) + 2 (x - a) (y ~ /3) cos <* = r Write an equation representing all circles with ^ the radius 5 whose centers lie on the X-axis ; on the F-axis. 11. Find the number of values of k for which the locus of (a) x 2 + y kx - 2 y + 5 k = 0, (b) x* + y* + 4'fce - 2y - k = 0, (c) x2 + y kx - 2 y + 4 fc = is a point-circle. ^4ns. (a) two; (b) none ; (c) one. 12. Plot the circles z 2 -f y x - 9 = 0, z 2 + y 2-4 x - 9 = 0, and x z + y + 4x -9 + k(x* + y*-4x- 9) = for fc=l, 3, 1, -5,. Must any values of k be excluded? 13. Plot the circles x 2 + y x = 0, x 2 + y 2-4 x = 0, and x 2 + y fc(z 2 + y2 4x) = for the values of k in problem 12. Must any values of k be excluded? 14. Plot the circles x 2 + y x + 9 = 0, x 2 + y 2-4 x + 9 = 0, and x 2 +?/ 2 + 4x fc(z 2 + y 2-4x + 9) = for k = - 3, -, - 5, -, ~~ 1 ~~ f ' ~ 1< What values of A; must be excluded? 54. Systems of circles. An equation of the form x 2 + if + Zta + y + F = will define a system of circles if one or more of the coefficients contain an arbitrary constant. Thus the equation ic 2 + ^2 _ r 2 = o represents the system of concentric circles whose centers are at the origin. Very interesting systems of circles, and the only systems we shall consider, are represented by equations analogous to (XIII), p. 110.

7 ' THE CIRCLE 121 II. Given two circles, theorem C l :x* + y* + D^x + E 1y + F l = Q Cz :x* + if + D 2 x + EJJ + F 2 \ the locus of the equation (II) x 2 + y* + D& + EM + J\ = Oj is a circle except when k = l. In this case the locus is a straight line. Proof. Clearing the parenthesis in (II) and collecting like terms in x and y, we obtain Dividing by 1-4- k we have The locus of this equation is a circle (Theorem I, p. 116). however, k = 1, we cannot divide by 1 + k. But in this case equation (II) becomes (A - D 2 )aj + (E l -E 2 )y+ (F, - F) = 0, which is of the first degree in x and y. Its locus is. then a straight line called the radical axis of C\ and C 2. Q.E.D. If, Corollary I. The center of the circle (II) lies upon the line joining the centers of C x and C 2 and divides that line into segments whose ratio is equal to k. For by Theorem I (p. 116) the center of C\ is PI - ( ~t - ^ and ) of Cz ia - \ *J -j / --- Pz(- ) The point dividing - P\P>2 into segments whose ratio equals Tc r-t + *(-^) -f +Kis (Theorem VII, p. 32) the -, point ^ 1 ~r" ft^ 1 "T~ A/ - simplifying, ( - ff ~ f 2 )' wllteh is th CeDter f (H) '

8 122 ANALYTIC GEOMETRY Corollary II. The equation of the radical axis of C and C 2 is (D, -D 2)x + (E, -Ez)y + (F, - F 2 ) = 0. Corollary III. The radical axis of two circles is perpendicular to the line joining their centers. Hint. Find the line joining the centers of C t and C 2 (Theorem VII, p. 88) and show that it is perpendicular to the radical axis hy Corollary III, p. 78. The system (II) may have three distinct forms, as illustrated in the following examples. These three forms correspond to the relative positions of Ci and C which 2, may intersect in two points, be tangent to each other, or not meet at all. Ex. 1. Plot the system of circles represented by x 2 + y 2 + 8x k(x* + y 2-4x - 9) = 0. Solution. The figure shows the circles z 2 + y 2 -f8z-9=:0 and z 2 + 2/2-4z-9 = plotted in heavy lines and the circles corresponding to k = 2, 5, 1, i, - 4, - f, and - ; these circles all pass through the intersection of the first two. The radical axis of the two circles plotted in heavy lines, which corresponds to k = 1, is the Y-axis.

9 THE CIRCLE 123 Ex. 2. Plot the system of circles represented by x 2 + y 2 + 8x + k(x 2 + y2-4x) = 0. X Solution. The figure shows the circles X 2 _j_ y i + 8 x = and x 2 -f y 2-4 x = plotted in heavy lines and the circles corresponding to k = 2, 3,, 5, 1, i, - 7, i, - 4, - 3, and - \. These circles are all tangent to the given circles at their point of tangency. The locus for k = 2 is the origin. Ex. 3. Plot the system of circles represented by x 2 + y 2 - lox k (x2 + y 2 -f 8x + 9) = 0. Solution. The figure shows the circles x 2 -f y 2 10 x + 9 = and x 2 + y' x + 9 = plotted in heavy lines and the circles corresponding to * = it 17, i, - 10, - TV, and - -^.

10 124 ANALYTIC GEOMETRY These circles all cut the dotted circle at right angles (problem 7). For k= - the locus is the point-circle (3, 0), and for k = 8 it is the point-circle ( 3, 0). In all three examples the radical axis, for which k = 1, is the F-axis. PROBLEMS 1. If Ci and C 2 intersect in PI and P 2, the system (II) consists of all circles passing through PI and P If Ci and C 2 are tangent, the system (II) consists of all circles tangent to Ci and <7 2 at their point of tangency. 3. The radical axis of two intersecting circles is their common chord, and of two tangent circles is the common tangent at their point of tangency. 4. Find the equation of the circle passing through the intersections of the circles x 2 + y 2 1 = and x 2 + y x = which passes through the point (3, 2). Ans. 7 x y 2-24 x - 19 = Two circles x 2 + y 2 + DIX + E^y + FI = and x 2 + y 2 + D 2 z + E zy 4>F2 = intersect at right angles when and only when DiD2 + EiE2-2 FI - 2 F 2 = 0. Hint. Construct a triangle by drawing the line of centers and the radii to a point of intersection I\. 6. The equation of the system (II) may be written in the form x2 + y 2 + k'x + F = 0, where Fis constant and k' arbitrary, if the axes of x arid y be respectively chosen as the line of centers and the radical axis of Ci and C The system in problem C consists of all circles whose intercepts on the F-axis are V F if F<0, which are tangent to the F-axis at the origin if F = 0, and which intersect the circle x 2 + y 2 = F at right angles if P>0. 8. The square of the length of the tangent from PI (1,2/1) to the circle X 2 + y 2 + DX + Ey + F = is Xi 2 + y^ + Dx l + Ey + F. Hint. Construct a right triangle by joining P t and the point of tangency to the center. 9. The locus of points from which tangents to two circles are equal is their radical axis. 10. Find the radical axes of the circles x- + y 2-4x = 0, x 2 + y 2 + 6x 8 y ~ 0, and x 2 + y' 2 + G x - 8 = taken by pairs, and show that they meet in a point. 11. Show that the radical axes of any three circles taken by pairs meet in' a point. 12. By means of problem 11 show that a circle may be drawn cutting any three circles at right angles. 13. Show that the radical axis of any pair of circles in the system (II) is the same as the radical axis of Oi and C How may problem 11 be stated if the three circles are point-circles?

CHAPTER III THE CURVE AND THE EQUATION. The bisectors of the adjacent angles formed by two lines is. To solve any locus problem involves two things :

CHAPTER III THE CURVE AND THE EQUATION 24. Locus of a point satisfying a given condition. The curve* (or group of curves) passing through all points which satisfy a given condition, and through no other