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Q: A circle has equation x2+y2-8x+2y-1=0 what is the radius of this circle
A: A circle has equation x^2+y^2-8x+2y-1=0, the radius of this circle is 3 sqrt 2. x^2+y^2-8x+2y-1=0 x^2-8x+16+y^2+2y+1=1+16+1 (x-4)^2+(y+1)^2=18 The center of the circle is (4, -1) and the radius is sqrt 18 = 3 sqrt 2.
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User: A circle has equation x2+y2-8x+2y-1=0 what is the radius of this circle

Weegy: I don't know!
Expert answered|truchocolate08|Points 110|

Question
Asked 4/6/2009 12:35:29 PM
Updated 7/23/2014 3:15:01 AM
1 Answer/Comment
This conversation has been flagged as incorrect.
Flagged by yeswey [7/23/2014 3:14:52 AM], Edited by andrewpallarca [7/23/2014 3:22:57 AM]
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A circle has equation x^2+y^2-8x+2y-1=0, the radius of this circle is 3 sqrt 2.
x^2+y^2-8x+2y-1=0
x^2-8x+16+y^2+2y+1=1+16+1
(x-4)^2+(y+1)^2=18
The center of the circle is (4, -1) and the radius is sqrt 18 = 3 sqrt 2.
Added 7/23/2014 3:15:01 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [7/23/2014 3:23:06 AM]
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