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