A circle is described by the equation x squared + y squared + 14 x + 2 y + 14 = 0 what are the coordinates for the center of the circle and length of the radius

(x-h) ^2 + (x-k) ^2=r^2

x^2+14x+___+y^2+2y+__ = - 14 (14/2) ^2 = 49, (2/2) ^2=1

x^2+14x+49+y^2+2y+1 = - 14+49+1

(x+7) ^2 + (x+1) ^2 = 36

Center = (-7, - 1) Radius = square root of 36 so 6

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