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a circle has equation x2+y2-8x+2y-1=0 what is the radius of this circle
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Asked 4/6/2009 12:36:48 PM
Updated 7/23/2014 3:14:02 AM
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Flagged by yeswey [7/23/2014 3:11:57 AM]
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User: a circle has equation x2+y2-8x+2y-1=0 what is the radius of this circle

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Expert answered|jubsona|Points 10|

User: what

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Asked 4/6/2009 12:36:48 PM
Updated 7/23/2014 3:14:02 AM
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Flagged by yeswey [7/23/2014 3:11:57 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.
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Added 7/23/2014 3:14:02 AM
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A circle has equation x2+y2-8x+2y-1=0 what is the radius of this circle
Weegy: I don't know! (More)
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Expert Answered
Updated 7/23/2014 3:15:01 AM
1 Answer/Comment
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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