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September 1st, 2017, 10:49 AM   #1
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Smile 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 11:05 AM.
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September 1st, 2017, 10:53 AM   #2
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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.
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September 1st, 2017, 11:06 AM   #3
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It's a single point. I've moved this problem to Geometry.
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September 1st, 2017, 11:09 AM   #4
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Irrelevant after moved. Delete button in Edit doesn't seem to work

Last edited by zylo; September 1st, 2017 at 11:11 AM.
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September 2nd, 2017, 05:30 AM   #5
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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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