Find the center and radius of the circle x^2-14x+y^2+4y=-28

We'll have to re-write the equation in the standard
form:

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

We'll have to complete the squares and we'll start
with:

x^2 - 14x + b^2

We'll
put x^2 = a^2 => x = a

-14x = -2*x*b => b =
14/2 => b = 7

b^2 =
49

To complete the square, we'll add and subtract
49:

x^2 - 14x + 49 - 49 = (x - 7)^2 - 49
(1)

y^2 + 4y + b^2

a^2 = y^2
=> y = a

4y = 2*y*b => b =
4/2

b = 2 => b^2 = 4

To
complete the square, we'll add and subtract 4:

y^2 + 4y + 4
- 4 = (y+2)^2 - 4 (2)

We'll add (1) +
(2):

(x - 7)^2 - 49 + (y+2)^2 - 4 =
-28

We'll move the constants to the right
side:

(x - 7)^2 + (y+2)^2 = -28 + 4 +
49

(x - 7)^2 + (y+2)^2 =
25

The center of the circle is C(7 , -2) and
the radius is r = 5.