Question

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).

Answer 1

When you complete the square:

x^2 + 6x + y^2 + 4y − 3 = 0
(x^2 + 6x) + (y^2 + 4y) = 3
(x^2 + 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4
(x + 3)^2 + (y + 2)^2 = 16 = 4^2

So the center is (−3, −2). The correct choice is:
(x + 3)^2 + (y + 2)^2 = 4^2, so the center is (−3, −2).