The point of tangency of the circle with center at (3,-2) and the line 3x + 4y = 26. A. ( 6,2) В. ( -6,2) С. (-6,- 2) D. (6,-2)

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The point of tangency of the circle with center at (3,-2) and the line 3x + 4y = 26.
A. (6,2)
В. ( -6,2)
С. ( -6,—2)
D. (6,-2)
The equation of the circle with center at (2,2) and passing through (4,5).
А. х2 + у2 — 4х - 4у + 5 %3D 0
В. х2 + у? — 4х — 4у — 5 3D 0
C. x? + y2 + 4x + 4y + 5 = 0
D. x2 + y2 + 4x + 4y – 5 = 0
Find the center of the circle passing through the points (-18 , -5 ) , (-7, -16 ) and
(4 , -5).
A. (7,5)
В. (-7,5)
С. ( -7,-5)
D. (7,-5)
The center of the circle x2 + y2 – 24x + 44 = 0
А. (12,0)
В. (-12,0)
С. (0,12)
D. (0,–12)
The center of the circle x2 + y2 – 24x + 44 = 0
А. (12,0)
В. (-12,0)
С. (0,12)
D. (0,–12)
The equation of the circle with center at the first quadrant and tangent to the lines
х %3D 8,у %3D 3and x —D 14.
A. (x – 11)2 + (y – 6)² = 9
В. (х — 6)? + (у — 11)? %3D 9
C. (x – 11)² + (y – 3)2 = 9
D. (x – 3)2 + (y – 11)? = 9
The equation of the circle with center at the first quadrant and tangent to the lines
x = 8, y = 3 and x = 14.
A. (x – 11)? + (y – 6)? = 9
В. (х — 6)2 + (у — 11)? %3D 9
С. (х — 11)2 + (у — 3)2 %3D 9
D. (x — 3)2 + (у - 11)2 %3D 9
Transcribed Image Text:The point of tangency of the circle with center at (3,-2) and the line 3x + 4y = 26. A. (6,2) В. ( -6,2) С. ( -6,—2) D. (6,-2) The equation of the circle with center at (2,2) and passing through (4,5). А. х2 + у2 — 4х - 4у + 5 %3D 0 В. х2 + у? — 4х — 4у — 5 3D 0 C. x? + y2 + 4x + 4y + 5 = 0 D. x2 + y2 + 4x + 4y – 5 = 0 Find the center of the circle passing through the points (-18 , -5 ) , (-7, -16 ) and (4 , -5). A. (7,5) В. (-7,5) С. ( -7,-5) D. (7,-5) The center of the circle x2 + y2 – 24x + 44 = 0 А. (12,0) В. (-12,0) С. (0,12) D. (0,–12) The center of the circle x2 + y2 – 24x + 44 = 0 А. (12,0) В. (-12,0) С. (0,12) D. (0,–12) The equation of the circle with center at the first quadrant and tangent to the lines х %3D 8,у %3D 3and x —D 14. A. (x – 11)2 + (y – 6)² = 9 В. (х — 6)? + (у — 11)? %3D 9 C. (x – 11)² + (y – 3)2 = 9 D. (x – 3)2 + (y – 11)? = 9 The equation of the circle with center at the first quadrant and tangent to the lines x = 8, y = 3 and x = 14. A. (x – 11)? + (y – 6)? = 9 В. (х — 6)2 + (у — 11)? %3D 9 С. (х — 11)2 + (у — 3)2 %3D 9 D. (x — 3)2 + (у - 11)2 %3D 9
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