The standard equation of any circle is:
(x-h)^2 + (y-k)^2 = r^2 where:
r is the radius and (h,k) are the coordinates of the center of the circle.
Now, the given equation is:
x^2 + y^2 + 14x + 2y + 14 = 0
We will try to use the given equation to reach the standard form of the equation of circle as follows:
(x^2 + 14x) + (y^2 + 2y) + 14 = 0
(x^2+2(7)(x) +7^2) - 7^2 + ( y^2+(2)(1)y+1^2) -1^2 + 14 = 0
(x+7)^2 +(y+1)^2 = 36 = (6)^2
Comparing the equation we reached to the standard form, we will find that:
-h = 7 .............> h=-7
-k = 1 ..............> k=-1
r^2 = 36 .............> r = 6
Therefore:
The center is found at: (-7 , -1)
radius = 6
