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).
