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 12:35:29 PM

Updated 7/23/2014 3:15:01 AM

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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]

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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3

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.

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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