Question

Using the following equation, find the center and radius of the circle by completing the square.

x2 + y2 + 6x − 6y + 2 = 0

center: (−3, 3), r = 4
center: (3, −3) r = 4
center: (3, −3), r = 16
center: (−3, 3), r = 16

Answer 1

x2 + y2 + 6x - 6y + 2 = 0
To complete square to a quadratic equation in its standard form we have:
ax2 + bx + c
Completing squares:
P (x) = (x + b / 2) ^ 2 - b ^ 2/4 + c
Keeping this in mind, we can complete square then:
x2 + y2 + 6x - 6y = -2
(x2 + 6x) + (y2 - 6y) = -2
((x + b / 2) ^ 2 - b ^ 2/4 + c) + ((y + b / 2) ^ 2 - b ^ 2/4 + c) = -2
((x + 6/2) ^ 2 - 6 ^ 2/4 + 0) + ((y + (-6) / 2) ^ 2 - (-6) ^ 2/4 + 0) = -2
((x + 3) ^ 2 - 9) + ((y - 3) ^ 2 - 9) = -2
((x + 3) ^ 2) + ((y - 3) ^ 2) - 9 - 9 = -2
((x + 3) ^ 2) + ((y - 3) ^ 2) - 18 = -2
((x + 3) ^ 2) + ((y - 3) ^ 2) = -2 + 18
((x + 3) ^ 2) + ((y - 3) ^ 2) = 16
((x + 3) ^ 2) + ((y - 3) ^ 2) = 4 ^ 2
Answer:
center: (-3, 3), r = 4