A circle is a set of points on the plane defined by the distance of any of them to a point #C (a, b)# given is constant. At this point #C# we will call it center of the circumference and the distance of this one to the points of the circumference receives the name of radius.

If a point #P (x, y)# of the plane belongs to the circumference, its distance from the center #C (a, b)# will be equal to the radius of the circle, #r#. The distance between #P# and #C# will be given by the module of #vec {CP}#:

#d (P, C) = |vec {CP}| = sqrt {(x - a)^2 + (y - b)^2}#,

which means that we will have:

#sqrt {(x - a)^2 + (y - b)^2} = r#.

If we raise everything to the square (to eliminate the root) and we develop the squares to eliminate the parentheses, we obtain:

#(sqrt {(x - a)^2 + (y - b)^2})^2 = r^2 rArr (x - a)^2 + (y - b)^2 = r^2#

#x^2 - 2 a x + a^2 + y^2 - 2 b y + b^2 = r^2#.

Rearranging terms:

#x^2 + y^2 - 2 a x - 2 b y + a^2 + b^2 - r^2 = 0#,

we obtain a generic formula for any circumference, which we can write of the form:

#x^2 + y^2 + m x + n y + p = 0#,

being:

#m = - 2 a#,

#n = - 2 b#, and

#p = a^2 + b^2 - r^2#.

So, if we write the given equation in the following way:

#2 x^2 + 2 y^2 - 8 x + 12 y + 2 = 0 rArr x^2 + y^2 - 4 x + 6 y + 1 = 0#

(dividing everything by 2), it is clear that we have the equation of a circumference with:

#m = - 4; n = 6; and p = 1#.

Given that:

#- 2 a = - 4 rArr a = 2#,

#- 2 b = 6 rArr b = - 3#, and

#a^2 + b^2 - r^2 = 1 rArr (2)^2 + (- 3)^2 - r^2 = 1 rArr r = 2 sqrt 3#,

it is clear that the proposed equation corresponds to a circumference of radius #r = 2 sqrt 3# and center at the point #(2, -3)#.