My Math Forum show that the equation x^2+y^2+4x+4=0 is a circle and determine its centre and radius

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 September 1st, 2017, 09:49 AM #1 Newbie   Joined: Aug 2017 From: Luweero, Uganda Posts: 4 Thanks: 0 Math Focus: differentiation show that the equation x^2+y^2+4x+4=0 is a circle and determine its centre and radius How can I Show that the equation x^2+y^2+4x+4=0 is a circle and determine its centre and radius? Last edited by skipjack; September 1st, 2017 at 10:05 AM.
 September 1st, 2017, 09:53 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Perform a substitutions for $x$ and $y$ that result in the form $u^2 + v^2 = r^$. Are you sure your equation is correct? It looks like an odd sort of circle to me.
 September 1st, 2017, 10:06 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,826 Thanks: 2160 It's a single point. I've moved this problem to Geometry.
 September 1st, 2017, 10:09 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Irrelevant after moved. Delete button in Edit doesn't seem to work Last edited by zylo; September 1st, 2017 at 10:11 AM.
 September 2nd, 2017, 04:30 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Complete the square in x: $\displaystyle x^2+ 4x= x^2+ 4x+ 4- 4= (x- 2)^2- 4$ (The graph of $\displaystyle x^2+ y^2+ 4x+ 4= 0$ is NOT a circle. It is the single point (-2, 0).)

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