Question

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

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