a p p e n d i g DISTANCE, CIRCLES, AND QUADRATIC EQUATIONS DISTANCE BETWEEN TWO POINTS IN THE PLANE Suppose that we are interested in finding the distance d between two points P (, ) and P (, ) in the -plane. If, as in Figure G., we form a right triangle with P and P as vertices, then it follows from Theorem B.4 in Appendi B that the sides of that triangle have lengths and. Thus, it follows from the Theorem of Pthagoras that d = + = ( ) + ( ) and hence we have the following result. G. theorem. The distance d between two points P (, ) and P (, ) in a coordinate plane is given b d = ( ) + ( ) () P (, ) d P (, ) Figure G. To appl Formula () the scales on the coordinate aes must be the same; otherwise, we would not have been able to use the Theorem of Pthagoras in the derivation. Moreover, when using Formula () it doesnot matter which point islabeled P and which one islabeled P, since reversing the pointschangesthe signsof and ; thishasno effect on the value of d because these quantities are squared in the formula. When it is important to emphasize the points, the distance between P and P isdenoted b d(p,p ) or d(p,p ). Eample Find the distance between the points (, 3) and (, 7). Solution. If we let (, ) be (, 3) and let (, ) be (, 7), then () ields A6 d = [ ( )] +[7 3] = 3 + 4 = 5 = 5
Appendi G: Distance, Circles, and Quadratic Equations A63 A(4, 6) B(, 3) Figure G. C(7, 5) Eample It can be shown that the converse of the Theorem of Pthagoras is true; that is, if the sides of a triangle satisf the relationship a + b = c, then the triangle must be a right triangle. Use this result to show that the points A(4, 6), B(, 3), and C(7, 5) are vertices of a right triangle. Solution. The points and the triangle are shown in Figure G.. From (), the lengths of the sides of the triangles are Since d(a,b) = ( 4) + ( 3 6) = 9 + 8 = 90 d(a,c) = (7 4) + (5 6) = 9 + = 0 d(b,c) = (7 ) +[5 ( 3)] = 36 + 64 = 00 = 0 [d(a,b)] +[d(a,c)] =[d(b,c)] it follows that ABC is a right triangle with hpotenuse BC. a Figure G.3 Figure G.4 (b a) P (, ) b a M(, ) b P (, ) THE MIDPOINT FORMULA It is often necessar to find the coordinates of the midpoint of a line segment joining two points in the plane. To derive the midpoint formula, we will start with two points on a coordinate line. If we assume that the points have coordinates a and b and that a b, then, as shown in Figure G.3, the distance between a and b is b a, and the coordinate of the midpoint between a and b is a + (b a) = a + b = (a + b) which is the arithmetic average of a and b. Had the points been labeled with b a, the same formula would have resulted (verif). Therefore, the midpoint of two points on a coordinate line is the arithmetic average of their coordinates, regardless of their relative positions. If we now let P (, ) and P (, ) be an two points in the plane and M(,) the midpoint of the line segment joining them (Figure G.4), then it can be shown using similar triangles that is the midpoint of and on the -ais and is the midpoint of and on the -ais, so = ( + ) and = ( + ) Thus, we have the following result. G. theorem (The Midpoint Formula). The midpoint of the line segment joining two points (, ) and (, ) in a coordinate plane is ( ( + ), ( + ) ) () r ( 0, 0 ) (, ) Eample 3 Find the midpoint of the line segment joining (3, 4) and (7, ). Solution. From () the midpoint is ( (3 + 7), ( 4 + )) = (5, ) Figure G.5 CIRCLES If ( 0, 0 ) is a fied point in the plane, then the circle of radius r centered at ( 0, 0 ) is the set of all points in the plane whose distance from ( 0, 0 ) is r (Figure G.5). Thus, a point
A64 Appendi G: Distance, Circles, and Quadratic Equations (, ) will lie on this circle if and onl if ( 0 ) + ( 0 ) = r or equivalentl, This is called the standard form of the equation of a circle. ( 0 ) + ( 0 ) = r (3) Eample 4 Find an equation for the circle of radius 4 centered at ( 5, 3). Solution. From (3) with 0 = 5, 0 = 3, and r = 4 we obtain ( + 5) + ( 3) = 6 If desired, this equation can be written in an epanded form b squaring the terms and then simplifing: ( + 0 + 5) + ( 6 + 9) 6 = 0 + + 0 6 + 8 = 0 Eample 5 (4, ). Find an equation for the circle with center (, ) that passes through Solution. The radius r of the circle is the distance between (4, ) and (, ),so r = ( 4) + ( ) = 5 We now know the center and radius, so we can use (3) to obtain the equation ( ) + ( + ) = 5 or + + 4 0 = 0 FINDING THE CENTER AND RADIUS OF A CIRCLE When ou encounter an equation of form (3), ou will know immediatel that its graph is a circle; its center and radius can then be found from the constants that appear in the equation: ( 0 ) + ( 0 ) = r -coordinate of the center is 0 -coordinate of the center is 0 radius squared + = Eample 6 equation of a circle center ( 0, 0 ) radius r ( ) + ( 5) = 9 ( + 7) + ( + ) = 6 + = 5 ( 4) + = 5 (, 5) ( 7, ) (0, 0) (4, 0) 3 4 5 5 Figure G.6 The unit circle The circle + =, which is centered at the origin and has radius, is of special importance; it is called the unit circle (Figure G.6). OTHER FORMS FOR THE EQUATION OF A CIRCLE An alternative version of Equation (3) can be obtained b squaring the terms and simplifing. This ields an equation of the form + + d + e + f = 0 (4) where d, e, and f are constants. (See the final equations in Eamples 4 and 5.)
Appendi G: Distance, Circles, and Quadratic Equations A65 Still another version of the equation of a circle can be obtained b multipling both sides of (4) b a nonzero constant A. This ields an equation of the form A + A + D + E + F = 0 (5) where A, D, E, and F are constants and A = 0. If the equation of a circle is given b (4) or (5), then the center and radius can be found b first rewriting the equation in standard form, then reading off the center and radius from that equation. The following eample shows how to do this using the technique of completing the square. In preparation for the eample, recall that completing the square is a method for rewriting an epression of the form + b as a difference of two squares. The procedure is to take half the coefficient of, square it, and then add and subtract that result from the original epression to obtain + b = + b + (b/) (b/) =[ + (b/)] (b/) Eample 7 Find the center and radius of the circle with equation (a) + 8 + + 8 = 0 (b) + + 4 8 = 0 Solution (a). right side: First, group the -terms, group the -terms, and take the constant to the ( 8) + ( + ) = 8 Net we want to add the appropriate constant within each set of parentheses to complete the square, and subtract the same constant outside the parentheses to maintain equalit. The appropriate constant is obtained b taking half the coefficient of the first-degree term and squaring it. This ields ( 8 + 6) 6 + ( + + ) = 8 from which we obtain ( 4) + ( + ) = 8 + 6 + or ( 4) + ( + ) = 9 Thus from (3), the circle has center (4, ) and radius 3. Solution (b). The given equation is of form (5) with A =. We will first divide through b (the coefficient of the squared terms) to reduce the equation to form (4). Then we will proceed as in part (a) of this eample. The computations are as follows: + + 8 = 0 We divided through b. ( + ) + = 8 ( + + 36) + = 8 ( + 6) + = 53 53 From (3), the circle has center ( 6, 0) and radius. + 36 We completed the square. DEGENERATE CASES OF A CIRCLE There is no guarantee that an equation of form (5) represents a circle. For eample, suppose that we divide both sides of (5) b A, then complete the squares to obtain ( 0 ) + ( 0 ) = k Depending on the value of k, the following situations occur:
A66 Appendi G: Distance, Circles, and Quadratic Equations (k > 0) The graph is a circle with center ( 0, 0 ) and radius k. (k = 0) The onl solution of the equation is = 0, = 0, so the graph is the single point ( 0, 0 ). (k < 0) The equation has no real solutions and consequentl no graph. Eample 8 Describe the graphs of (a) ( ) + ( + 4) = 9 (b) ( ) + ( + 4) = 0 Solution (a). There are no real values of and that will make the left side of the equation negative. Thus, the solution set of the equation is empt, and the equation has no graph. Solution (b). The onl values of and that will make the left side of the equation 0 are =, = 4. Thus, the graph of the equation is the single point (, 4). The following theorem summarizes our observations. The last two cases in Theorem G.3 are called degenerate cases. In spite of the fact that these degenerate cases can occur, (6) isoften called the general equation of a circle. G.3 theorem. An equation of the form A + A + D + E + F = 0 (6) where A = 0, represents a circle, or a point, or else has no graph. THE GRAPH of = a + b + c An equation of the form = a + b + c (a = 0) (7) is called a quadratic equation in. Depending on whether a is positive or negative, the graph, which is called a parabola, has one of the two forms shown in Figure G.7. In both cases the parabola is smmetric about a vertical line parallel to the -ais. This line of smmetr cuts the parabola at a point called the verte. The verte is the low point on the curve if a>0 and the high point if a<0. Verte b/(a) Verte b/(a) Figure G.7 = a + b + c a > 0 = a + b + c a < 0 In the eercises (Eercise 78) we will help the reader show that the -coordinate of the verte is given b the formula = b (8) a
Appendi G: Distance, Circles, and Quadratic Equations A67 0 3 = 3 With the aid of this formula, a reasonabl accurate graph of a quadratic equation in can be obtained b plotting the verte and two points on each side of it. Eample 9 Sketch the graph of (a) = (b) = + 4 5 Solution (a). The equation is of form (7) with a =, b =, and c =, so b (8) the -coordinate of the verte is = b a = Using this value and two additional values on each side, we obtain Figure G.8. Figure G.8 - - 3 4 0 3 4 Figure G.9 = = + 4 5 5 5-3 4 5 = + 4 5 Solution (b). The equation is of form (7) with a =, b = 4, and c = 5, so b (8) the -coordinate of the verte is = b a = Using this value and two additional values on each side, we obtain the table and graph in Figure G.9. Quite often the intercepts of a parabola = a + b + c are important to know. The -intercept, = c, results immediatel b setting = 0. However, in order to obtain the -intercepts, if an, we must set = 0 and then solve the resulting quadratic equation a + b + c = 0. Eample 0 Solve the inequalit > 0 Solution. Because the left side of the inequalit does not have readil discernible factors, the test-point method illustrated in Eample 4 of Appendi A is not convenient to use. Instead, we will give a graphical solution. The given inequalit is satisfied for those values of where the graph of = is above the -ais. From Figure G.8 those are the values of to the left of the smaller intercept or to the right of the larger intercept. To find these intercepts we set = 0 to obtain = 0 Solving b the quadratic formula gives = b ± b 4ac = ± = ± 3 a Thus, the -intercepts are = + 3.7 and = 3 0.7 and the solution set of the inequalit is (, 3) ( + 3, + ) Note that the decimal approimationsof the interceptscalculated in the preceding eample agree with the graph in Figure G.8. Observe, however, that we used the eact values of the intercepts to epressthe solution. The choice of eact versusapproimate valuesisoften a matter of judgment that dependson the purpose for which the valuesare to be used. Numerical approimationsoften provide a sense of size that eact values do not, but the can introduce severe errors if not used with care.
A68 Appendi G: Distance, Circles, and Quadratic Equations Eample From Figure G.9 we see that the parabola = + 4 5 has no -intercepts. This can also be seen algebraicall b solving for the -intercepts. Setting = 0 and solving the resulting equation s b the quadratic formula ields + 4 5 = 0 = 4 ± 6 0 = ± i 0 Because the solutions are not real numbers, there are no -intercepts. Distance (m) Earth surface s 40 30 0 0 3 4 5 6 Time (s) Figure G.0 t Eample A ball is thrown straight up from the surface of the Earth at time t = 0s with an initial velocit of 4.5 m/s. If air resistance is ignored, it can be shown that the distance s (in meters) of the ball above the ground after t seconds is given b s = 4.5t 4.9t (9) (a) Graph s versus t, making the t-ais horizontal and the s-ais vertical. (b) How high does the ball rise above the ground? Solution (a). Equation (9) is of form (7) with a = 4.9, b = 4.5, and c = 0, so b (8) the t-coordinate of the verte is t = b 4.5 = a ( 4.9) =.5 s and consequentl the s-coordinate of the verte is b/(a) Verte The factored form of (9) is s = 4.5(.5) 4.9(.5) = 30.65 m s = 4.9t(5 t) so the graph has t-intercepts t = 0 and t = 5. From the verte and the intercepts we obtain the graph shown in Figure G.0. = a + b + c a > 0 Solution (b). From the s-coordinate of the verte we deduce that the ball rises 30.65 m above the ground. b/(a) Verte THE GRAPH of = a + b + c If and are interchanged in (7), the resulting equation, = a + b + c = a + b + c a < 0 is called a quadratic equation in. The graph of such an equation is a parabola with its line of smmetr parallel to the -ais and its verte at the point with -coordinate = b/(a) (Figure G.). Some problems relating to such equations appear in the eercises. Figure G.
Appendi G: Distance, Circles, and Quadratic Equations A69 EXERCISE SET G. Where in this section did we use the fact that the same scale was used on both coordinate aes? 5 Find (a) the distance between A and B (b) the midpoint of the line segment joining A and B.. A(, 5), B(, ) 3. A(7, ), B(, 9) 4. A(, 0), B( 3, 6) 5. A(, 6), B( 7, 4) 6 0 Use the distance formula to solve the given problem. 6. Prove that (, ), (, 8), and (4, 0) lie on a straight line. 7. Prove that the triangle with vertices (5, ), (6, 5), (, ) is isosceles. 8. Prove that (, 3), (4, ), and (, 6) are vertices of a right triangle and then specif the verte at which the right angle occurs. 9. Prove that (0, ), ( 4, 8), and (3, ) lie on a circle with center (, 3). 0. Prove that for all values of t the point (t, t 6) is equidistant from (0, 4) and (8, 0).. Find k, given that (,k)is equidistant from (3, 7) and (9, ).. Find and if (4, 5) is the midpoint of the line segment joining ( 3, ) and (, ). 3 4 Find an equation of the given line. 3. The line is the perpendicular bisector of the line segment joining (, 8) and ( 4, 6). 4. The line is the perpendicular bisector of the line segment joining (5, ) and (4, 8). 5. Find the point on the line 4 + 3 = 0 that is equidistant from (3, 3) and (7, 3). [Hint: First find an equation of the line that is the perpendicular bisector of the line segment joining (3, 3) and (7, 3).] 6. Find the distance from the point (3, ) to the line (a) = 4 (b) =. 7. Find the distance from (, ) to the line 4 3 + 0 = 0. [Hint: Find the foot of the perpendicular dropped from the point to the line.] 8. Find the distance from (8, 4) to the line 5 + 36 = 0. [Hint: See the hint in Eercise 7.] 9. Use the method described in Eercise 7 to prove that the distance d from ( 0, 0 ) to the line A + B + C = 0is d = A 0 + B 0 + C A + B 0. Use the formula in Eercise 9 to solve Eercise 7.. Use the formula in Eercise 9 to solve Eercise 8.. Prove: For an triangle, the perpendicular bisectors of the sides meet at a point. [Hint: Position the triangle with one verte on the -ais and the opposite side on the -ais, so that the vertices are (0,a), (b, 0), and (c, 0).] 3 4 Find the center and radius of each circle. 3. (a) + = 5 (b) ( ) + ( 4) = 6 (c) ( + ) + ( + 3) = 5 (d) + ( + ) = 4. (a) + = 9 (b) ( 3) + ( 5) = 36 (c) ( + 4) + ( + ) = 8 (d) ( + ) + = 5 3 Find the standard equation of the circle satisfing the given conditions. 5. Center (3, ); radius = 4. 6. Center (, 0); diameter = 8. 7. Center ( 4, 8); circle is tangent to the -ais. 8. Center (5, 8); circle is tangent to the -ais. 9. Center ( 3, 4); circle passes through the origin. 30. Center (4, 5); circle passes through (, 3). 3. A diameter has endpoints (, 0) and (0, ). 3. A diameter has endpoints (6, ) and (, 3). 33 44 Determine whether the equation represents a circle, a point, or no graph. If the equation represents a circle, find the center and radius. 33. + 4 = 0 34. + + 8 + 8 = 0 35. + + 4 4 = 0 36. 6 + 6 6 + 6 = 3 37. + + + + = 0 38. + 4 6 + 3 = 0 39. 9 + 9 = 40. ( /4) + ( /4) = 4. + + 0 + 6 = 0 4. + 0 + 9 = 0 43. 6 + 6 + 40 + 6 7 = 0 44. 4 + 4 6 4 = 9 45. Find an equation of (a) the bottom half of the circle + = 6 (b) the top half of the circle + + 4 + = 0. 46. Find an equation of (a) the right half of the circle + = 9 (b) the left half of the circle + 4 + 3 = 0. 47. Graph (a) = 5 (b) = 5 + 4.
A70 Appendi G: Distance, Circles, and Quadratic Equations 48. Graph (a) = 4 (b) = 3 + 4. 49. Find an equation of the line that is tangent to the circle + = 5 at the point (3, 4) on the circle. 50. Find an equation of the line that is tangent to the circle at the point P on the circle (a) + + = 9; P(, ) (b) + 6 + 4 = 3; P(4, 3). 5. For the circle + = 0 and the point P(, ): (a) Is P inside, outside, or on the circle? (b) Find the largest and smallest distances between P and points on the circle. 5. Follow the directions of Eercise 5 for the circle + 4 = 0 and the point P ( ) 3, 5. 53. Referring to the accompaning figure, find the coordinates of the points T and T, where the lines L and L are tangent to the circle of radius with center at the origin. L L T T (3, 0) Figure E-53 54. Apoint (, ) moves so that its distance to (, 0) is times its distance to (0, ). (a) Show that the point moves along a circle. (b) Find the center and radius. 55. A point (, ) moves so that the sum of the squares of its distances from (4, ) and (, 5) is 45. (a) Show that the point moves along a circle. (b) Find the center and radius. 56. Find all values of c for which the sstem of equations { = 0 ( c) + = has 0,,, 3, or 4 solutions. [Hint: Sketch a graph.] 57 70 Graph the parabola and label the coordinates of the verte and the intersections with the coordinate aes. 57. = + 58. = 3 59. = + 3 60. = 3 4 6. = + 4 + 5 6. = + 63. = ( ) 64. = (3 + ) 65. + = 0 66. + 8 + 8 = 0 67. = 3 + 68. = + + 69. = + + 70. = 4 + 5 7. Find an equation of (a) the right half of the parabola = 3 (b) the left half of the parabola =. 7. Find an equation of (a) the upper half of the parabola = 5 (b) the lower half of the parabola =. 73. Graph (a) = + 5 (b) = 4. 74. Graph (a) = + 4 (b) = 3 +. 75. If a ball is thrown straight up with an initial velocit of 3 ft/s, then after t seconds the distance s above its starting height, in feet, is given b s = 3t 6t. (a) Graph this equation in a ts-coordinate sstem (t-ais horizontal). (b) At what time t will the ball be at its highest point, and how high will it rise? 76. A rectangular field is to be enclosed with 500 ft of fencing along three sides and b a straight stream on the fourth side. Let be the length of each side perpendicular to the stream, and let be the length of the side parallel to the stream. (a) Epress in terms of. (b) Epress the area A of the field in terms of. (c) What is the largest area that can be enclosed? 77. A rectangular plot of land is to be enclosed using two kinds of fencing. Two opposite sides will have heav-dut fencing costing $3/ft, and the other two sides will have standard fencing costing $/ft. A total of $600 is available for the fencing. Let be the length of each side with the heavdut fencing, and let be the length of each side with the standard fencing. (a) Epress in terms of. (b) Find a formula for the area A of the rectangular plot in terms of. (c) What is the largest area that can be enclosed? 78. (a) B completing the square, show that the quadratic equation = a + b + c can be rewritten as ( = a + b ) ) + (c b a 4a if a = 0. (b) Use the result in part (a) to show that the graph of the quadratic equation = a + b + c has its high point at = b/(a)if a<0 and its low point there if a>0. 79 80 Solve the given inequalit. 79. (a) + 5 < 0 (b) + 3 > 0 80. (a) + > 0 (b) 4 + 6 < 0 8. At time t = 0 a ball is thrown straight up from a height of 5 ft above the ground. After t seconds its distance s, in feet, above the ground is given b s = 5 + 40t 6t. (a) Find the maimum height of the ball above the ground.
Appendi G: Distance, Circles, and Quadratic Equations A7 (b) Find, to the nearest tenth of a second, the time when the ball strikes the ground. (c) Find, to the nearest tenth of a second, how long the ball will be more than ft above the ground. 8. Find all values of at which points on the parabola = lie below the line = + 3.