what is the solution for finding the radius and center of this equation

Question

x^2 + 14x + y^2 + 12y = – 49.

Answer 1

Given circle equation x ^2 + 14x + y ^2 + 12y  = -49.

Add 49 to each side.

x ^2 + y ^2 + 14x + 12y + 49  = 0

Compare it to circle equation x ^2 + y ^2 + 2gx + 2fy + c  = 0.

c = 49

2g = 14

Divide each side by 2.

g  = 7

2f  = 12

Divide each side by 2.

f  = 6

Center (-g, -f ) = ( -7 , -6 )

Radius(r ) = √(g ^2 + f ^2 - c )

= √(49 + 36 - 49)

= √ 36

= 6

Given circle center = ( -7 , -6 )

r = 6.

Answer 2

The equation is x² + 14x + y² + 12y = - 49.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here, x coefficient = 14, so, (half the x coefficient)² = (14/2)²= 49.

Here, y coefficient = 12, so, (half the y coefficient)² = (12/2)²= 36.

Add 49 and 36 to each side.

x² + 14x + 49 + y² + 12y + 36 = - 49 + 49 + 36

(x + 7)² + (y + 6)² = 36

(x - (- 7))² + (y -(- 6))² = 6² .

Compare the above equation with the standard form of a circle equation : ( x - h )² + ( y - k )² = r², where, (h, k) is the center of the circle, and r is the radius.

The center (h, k) = (- 7, - 6).

The radius (r) = 6 units.