4 November, 22:47

# 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 = 0x2 + 6x + y2 + 4y - 3 = 0(x2 + 6x) + (y2 + 4y) = 3(x2 + 6x + 9) + (y2 + 4y + 4) = 3 + 9 + 4Which 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).

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1. 4 November, 22:50
0
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.

x ² + 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)
2. 4 November, 22:54
0
The correct answer is D) (x+3) ² + (y+2) ² = 4²; so the center is (-3,-2)