Show that the following equation represents a circle for every $a$:$$x^2+y^2-2ax-14y+40=0 .$$
My answer is: \begin{align*} (x-a)^2 - a^2 + (y-7)^2 -49+40 & =0\\ (x-a)^2+(y-7)^2 & =a^2+9\\ \implies \quad a^2+9 & >0\\ a^2 & > -9 \end{align*}
For every $ a \in \mathbb{R}$, $a^2 > -9$, so the equation represents a circle for every $a$.
Is this correct? And especially, my conclusion, how would that be written correctly, because I know I've probably done something wrong?