20
Find the center and radius of the circle passing through the points (3, 8), (9,6), and (13, -2).
By Problem 4.9, the circle has an equation x
2 + y
2 + Dx + Ey + F = 0. Substituting the values (x, y) at
each of the three given points, we obtain three equations: 3D + 8E + F= -73, 9D + 6E + F= -117,
13D - 2E + F = —173. First, we eliminate F (subtracting the second equation from the first, and subtracting the
third equation from the first):
-10D + 10E = 100
or, more simply
-3D+ £ = 22
-D + E = W
Subtracting the second equation from the first, -2D = 12, or D = -6. So, E = 10 + D = 4, and F =
-73-3Z)-8£ = -87. Hence, we obtain x
2 + y
2 -6x + 4y - 87 = 0. Complete the square: (x - 3)
2 +
(y +2)
2 -87 = 9 + 4, or, (x - 3)
2 + (y + 2)
2 = 100. Hence, the center is (3,-2) and the radius is VT50 =
10. Answer
Use a geometrical/coordinate method to find the standard equation of the circle passing through the points
A(0, 6), B(12,2), and C(16, -2).
Find the perpendicular bisector of AB. The slope of AB is - j and therefore the slope of the perpendicular
bisector is 3. Since it passes through the midpoint (6, 4) of AB, a point-slope equation for it is y - 4 = 3(,x — 6),
or, equivalently, y = 3x — 14. Now find the perpendicular bisector of BC. A similar calculation yields
y = x - 14. Since the perpendicular bisector of a chord of a circle passes through the center of the circle, the
center of the circle will be the intersection point of y = 3x - 14 and y = x — 14. Setting 3x - 14 = x — 14,
we find x = 0. So, y = x - 14 = —14. Thus, the center is (0,—14). The radius is the distance between the
center (0,-14) and any point on the circle, say (0,6): V(° ~ °)
2 + (-14 - 6)
2 = V400 = 20. Hence, the
standard equation is x
2 + (y + 14)
2 = 400.
Find the graph of the equation 2x
2 + 2y
2 — x = 0.
This is the
First divide by 2: x
2 + y
2 - \x = 0, and then complete the square: (x - \)
2 + y
2 = \.
standard equation of the circle with center (?, 0) and radius \.
For what value(s) of k does the circle (x - k)
2 + (y - 2k)
2 = 10 pass through the point (1,1)?
(1 - k)
2 + (1 -2&)
2 = 10. Squaring out and simplifying, we obtain 5A:
2 - 6fc - 8 = 0. The left side factors
into (5Jt+ 4)()t-2). Hence, the solutions are & = -4/5 and k = 2.
Find the centers of the circles of radius 3 that are tangent to both the lines x = 4 and y = 6.
a, ft) be a center. The conditions of tangency imply that |a —4| = 3 and \b — 6|=3 (see Fig. 4-1).I Let (a, ft) be a center. The conditions of tangency imply that |a —4| = 3 and \b — 6|=3 (see Fig. 4-1).
Hence, a = l or a — I, and b = 3 or b = 9. Thus, there are four circles.
Fig. 4-1
4.16
Determine the value of k so that x
2 + y
2 — 8x + lOy + k = 0 is the equation of a circle of radius 7.
Complete the square: (x - 4)
2 + (y + 5)
2 + k = 16 + 25. Thus, (x - 4)
2 + (y + 5)
2 = 41 - k. So,
Find the standard equation of the circle which has as a diameter the segment joining the points (5, -1) and
(-3, 7). The center is the midpoint (1, 3) of the given segment. The radius is the distance between (1, 3) and
4.11
4.12
4.13
4.14
4.15
4.17
-6D+2E=44
Hence, the equation is
Squaring, 41-A: = 49, and, therefore, k=-8.
(5,-1):
(x-1)2+(y-3)2=32.
CHAPTER 4
Find the center and radius of the circle passing through the points (3, 8), (9,6), and (13, -2).
By Problem 4.9, the circle has an equation x
2 + y
2 + Dx + Ey + F = 0. Substituting the values (x, y) at
each of the three given points, we obtain three equations: 3D + 8E + F= -73, 9D + 6E + F= -117,
13D - 2E + F = —173. First, we eliminate F (subtracting the second equation from the first, and subtracting the
third equation from the first):
-10D + 10E = 100
or, more simply
-3D+ £ = 22
-D + E = W
Subtracting the second equation from the first, -2D = 12, or D = -6. So, E = 10 + D = 4, and F =
-73-3Z)-8£ = -87. Hence, we obtain x
2 + y
2 -6x + 4y - 87 = 0. Complete the square: (x - 3)
2 +
(y +2)
2 -87 = 9 + 4, or, (x - 3)
2 + (y + 2)
2 = 100. Hence, the center is (3,-2) and the radius is VT50 =
10. Answer
Use a geometrical/coordinate method to find the standard equation of the circle passing through the points
A(0, 6), B(12,2), and C(16, -2).
Find the perpendicular bisector of AB. The slope of AB is - j and therefore the slope of the perpendicular
bisector is 3. Since it passes through the midpoint (6, 4) of AB, a point-slope equation for it is y - 4 = 3(,x — 6),
or, equivalently, y = 3x — 14. Now find the perpendicular bisector of BC. A similar calculation yields
y = x - 14. Since the perpendicular bisector of a chord of a circle passes through the center of the circle, the
center of the circle will be the intersection point of y = 3x - 14 and y = x — 14. Setting 3x - 14 = x — 14,
we find x = 0. So, y = x - 14 = —14. Thus, the center is (0,—14). The radius is the distance between the
center (0,-14) and any point on the circle, say (0,6): V(° ~ °)
2 + (-14 - 6)
2 = V400 = 20. Hence, the
standard equation is x
2 + (y + 14)
2 = 400.
Find the graph of the equation 2x
2 + 2y
2 — x = 0.
This is the
First divide by 2: x
2 + y
2 - \x = 0, and then complete the square: (x - \)
2 + y
2 = \.
standard equation of the circle with center (?, 0) and radius \.
For what value(s) of k does the circle (x - k)
2 + (y - 2k)
2 = 10 pass through the point (1,1)?
(1 - k)
2 + (1 -2&)
2 = 10. Squaring out and simplifying, we obtain 5A:
2 - 6fc - 8 = 0. The left side factors
into (5Jt+ 4)()t-2). Hence, the solutions are & = -4/5 and k = 2.
Find the centers of the circles of radius 3 that are tangent to both the lines x = 4 and y = 6.
a, ft) be a center. The conditions of tangency imply that |a —4| = 3 and \b — 6|=3 (see Fig. 4-1).I Let (a, ft) be a center. The conditions of tangency imply that |a —4| = 3 and \b — 6|=3 (see Fig. 4-1).
Hence, a = l or a — I, and b = 3 or b = 9. Thus, there are four circles.
Fig. 4-1
4.16
Determine the value of k so that x
2 + y
2 — 8x + lOy + k = 0 is the equation of a circle of radius 7.
Complete the square: (x - 4)
2 + (y + 5)
2 + k = 16 + 25. Thus, (x - 4)
2 + (y + 5)
2 = 41 - k. So,
Find the standard equation of the circle which has as a diameter the segment joining the points (5, -1) and
(-3, 7). The center is the midpoint (1, 3) of the given segment. The radius is the distance between (1, 3) and
4.11
4.12
4.13
4.14
4.15
4.17
-6D+2E=44
Hence, the equation is
Squaring, 41-A: = 49, and, therefore, k=-8.
(5,-1):
(x-1)2+(y-3)2=32.
CHAPTER 4