3. Circumscribing the triangle with vertices (-1, -4),(3, -2), (5, 2).

College Algebra (MindTap Course List)
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Chapter7: Conic Sections And Quadratic Systems
Section7.1: The Circle And The Parabola
Problem 88E
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Exercise Question No. 3
Squaring both sides of equation (1) and
terms, we find the equation of the locus to be
-
It is a simple matter to show that, if k 1, then (2) 18 Ghe
and radius
equation of a circle with center at
k2
2ck
equal to the numerical value of
k2
If k = 1, equation
- 1
0, of the
(2) degenerates into the perpendicular bisector, x =
line segment joining A and B.
Note that, disassociating the result from the tools (coordi-
nate system, distance formula, etc.) used to obtain those
results, we, have shown that the locus in question is a circle
with its center on the line joining the fixed points.
Exercises
In Exs. 1-25, find the equation of the circle.
1. Through (2, 3), (3, 4), (–1, 2). Solve by two methods.
2. Through (3, 1), (5, 3), (-3, -1). Soive by two methods.
3. Circumscribing the triangle with vertices (-1, -4), (3, -2), (5, 2).
4. Circumscribing the triangle with vertices (a, 0), (a, a), (0, 2a).
Ans. z + y2+ ax- ay - 2a2 0.
5. Passing through the points (-1, <3), (-5, 3), and having its center
Ans. (t +6)2 + (y +2)3 = 26.
on the line z – 2y + 2 = 0.
6. Passing through the points (-4, -2), (2, 0), and having its center on
- 0.
Ans. z2 + y - 2x + 14y
7. Touching the line z+ 2y = 8 at (0, 4) and passing through (3, 7).
the line 5z -
2y 19.
Ans. (z- 1)2 + (y – 6)ª = 5.
8. Touching the line 4z-3y = 28 at (4, -4) and passing through
24 0.
%3D
(-3, –5,.
9. Touching the line 3z
Ans. z2 +y+ 2y
- 2y 5 at (3, 2) and passing through (-2, 1).
%3D
10. Touching the line z +y = 8 at (2, 6) and passing through (4, 0).
11. Touching the z-axis and passing through the points (3, 1), (10, 8).
Ans. Centers: (-2, 13), (6, 5).
Transcribed Image Text:Squaring both sides of equation (1) and terms, we find the equation of the locus to be - It is a simple matter to show that, if k 1, then (2) 18 Ghe and radius equation of a circle with center at k2 2ck equal to the numerical value of k2 If k = 1, equation - 1 0, of the (2) degenerates into the perpendicular bisector, x = line segment joining A and B. Note that, disassociating the result from the tools (coordi- nate system, distance formula, etc.) used to obtain those results, we, have shown that the locus in question is a circle with its center on the line joining the fixed points. Exercises In Exs. 1-25, find the equation of the circle. 1. Through (2, 3), (3, 4), (–1, 2). Solve by two methods. 2. Through (3, 1), (5, 3), (-3, -1). Soive by two methods. 3. Circumscribing the triangle with vertices (-1, -4), (3, -2), (5, 2). 4. Circumscribing the triangle with vertices (a, 0), (a, a), (0, 2a). Ans. z + y2+ ax- ay - 2a2 0. 5. Passing through the points (-1, <3), (-5, 3), and having its center Ans. (t +6)2 + (y +2)3 = 26. on the line z – 2y + 2 = 0. 6. Passing through the points (-4, -2), (2, 0), and having its center on - 0. Ans. z2 + y - 2x + 14y 7. Touching the line z+ 2y = 8 at (0, 4) and passing through (3, 7). the line 5z - 2y 19. Ans. (z- 1)2 + (y – 6)ª = 5. 8. Touching the line 4z-3y = 28 at (4, -4) and passing through 24 0. %3D (-3, –5,. 9. Touching the line 3z Ans. z2 +y+ 2y - 2y 5 at (3, 2) and passing through (-2, 1). %3D 10. Touching the line z +y = 8 at (2, 6) and passing through (4, 0). 11. Touching the z-axis and passing through the points (3, 1), (10, 8). Ans. Centers: (-2, 13), (6, 5).
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