Mrs. culland is finding the center of a circle whose equation is x2 + y2 + 6x + 4y – 3 = 0 by completing the square. her work is shown.

x2 + y2 + 6x + 4y – 3 = 0

x2 + 6x + y2 + 4y – 3 = 0

(x2 + 6x) + (y2 + 4y) = 3

(x2 + 6x + 9) + (y2 + 4y + 4) = 3 + 9 + 4

which completes the work correctly?

(x – 3)2 + (y – 2)2 = 42, so the center is (3, 2).
(x + 3)2 + (y + 2)2 = 42, so the center is (3, 2).
(x – 3)2 + (y – 2)2 = 42, so the center is (–3, –2).
(x + 3)2 + (y + 2)2 = 42, so the center is (–3, –2).

X² + y² + 6x + 4y – 3 = 0

Move the loose number over to the other side:
x² + y² + 6x + 4y = 3

Group the x-stuff together. Group the y-stuff together.
x² + 6x + y² + 4y = 3

Take the x-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation. Do the same with the y-term coefficient.  Convert the left side to squared form, and simplify the right side.

x-term coefficient: (6 * 1/2)² = 3² = 9
y-term coefficient: (4 * 1/2)² = 2² = 4

(x² + 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4

(x+3)² + (y+2)² = 16

x + 3 = 0 ; x = -3
y + 2 = 0 ; y = -2

(x+3)² + (y+2)² = 4² ; so the center is (-3,-2)

The correct answer is D) (x+3)² + (y+2)² = 4² ; so the center is (-3,-2)

Yes of course i will indeed you with your problem.