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$x^2+y^2-2ax-14y+40=0$

Show that the equation is a circle for every $a$.

My answer is:

$(x-a)^2 - a^2 + (y-7)^2 -49+40=0$

$(x-a)^2+(y-7)^2=a^2+9$

$a^2+9>0$

$a^2 > -9$

For every $ a \in \mathbb{R}$, $a^2 > -9$ , so the equation is a circle for every $a$.

Is this correct? And especially, my conclusion, how would that be written correctly, because I know I've probably did something wrong.

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2  
It seems all correct to me. –  Shahab Oct 14 '12 at 15:47
    
Even the notation of my conclusion? –  JohnHedge Oct 14 '12 at 15:48
    
I've slightly edited the conclusion. My advice is "When writing a formula, be able to translate it in your language": symbols are just a useful shortcut, they don't change the meaning of what you're saying. –  Andrea Orta Oct 14 '12 at 15:52
    
It would be better if you start from your equation and complete the squares to obtain the other equation. It is just a matter of formality. The part about $a^2$ is trivially true. That will provide the radius of the circle. –  Sigur Oct 14 '12 at 16:12
    
@AndreaOrta thank you –  JohnHedge Oct 14 '12 at 16:22

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