Question

By completing the square, calculate the center and radius of the circle given by the equation 3x2+3y2−12x+6y+12=0
.(1 point)

center: (6,−3)
; radius: 6

center: (12,−6)
; radius: 12

center: (4,−2)
; radius: 4

center: (2,−1)
; radius: 1

Answer 1

Factor 3: 3(x^2+y^2−4x+2y+4)=0 → x^2−4x+y^2+2y+4=0.

Complete the square:
(x^2−4x) + (y^2+2y) = (x−2)^2−4 + (y+1)^2−1.

So (x−2)^2 + (y+1)^2 −1 = 0 ⇒ (x−2)^2 + (y+1)^2 = 1.

Center: (2, −1); radius: 1.