What is the equation of a circle having (10,2) and (6,-4) as endpoints of its diameter? x² + y² 16x - 2y + 52 = 0 - x² + y²-16x + 2y + 52 = 0 Option 1 Option 2 x² + y² + 16x - 2y + 52 = 0 x² + y²-16x + 2y - 104 = 0 Option 3 Option 4
What is the equation of a circle having (10,2) and (6,-4) as endpoints of its diameter? x² + y² 16x - 2y + 52 = 0 - x² + y²-16x + 2y + 52 = 0 Option 1 Option 2 x² + y² + 16x - 2y + 52 = 0 x² + y²-16x + 2y - 104 = 0 Option 3 Option 4
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 69E
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![What is the equation of a circle having (10,2) and (6,-4) as endpoints of its
diameter?
x² + y²16x - 2y + 52 = 0
x² + y²16x + 2y + 52 = 0
Option 1
Option 2
x² + y² + 16x - 2y + 52 = 0
x² + y²-16x + 2y - 104 = 0
Option 3
Option 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a044a49-57b5-43ec-9961-a9e794cae8d8%2F04dd7894-e596-4de1-b866-6075bd1424cb%2F8nmfoph_processed.jpeg&w=3840&q=75)
Transcribed Image Text:What is the equation of a circle having (10,2) and (6,-4) as endpoints of its
diameter?
x² + y²16x - 2y + 52 = 0
x² + y²16x + 2y + 52 = 0
Option 1
Option 2
x² + y² + 16x - 2y + 52 = 0
x² + y²-16x + 2y - 104 = 0
Option 3
Option 4
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