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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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    @AndreaOrta thank yo$u$2012-10-14

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