Remember that the general form for a circle in the Cartesian co-ordinate system is:
$$(x-a)^{2}+(y-b)^{2}=r^{2},\tag{1}$$
Where the center of the circle is at $(a,b)$ and the radius is $r$.
Therefore, for your example $x^{2}-4x+y^{2}+6y=51$, we must show that it can be reduced to this form to show it is a circle.
We begin by completing the square for $x$, to give us:
$$x^{2}-4x=(x-2)^{2}-4,$$
Then, similarly for $y$:
$$y^{2}+6y=(y+3)^{2}-9,$$
Putting this in our original form, we have:
$$(x-2)^{2}-4+(y+3)^{2}-9=51$$
Re-arranging, we get:
$$(x-2)^{2}+(y+3)^{2}=64$$
Which by $(1)$, is a circle with center $(2,-3)$ and radius $\sqrt{64}=8$.
Hope this helps!