Asked by help shaikha

The center of circle x2+ y2+ 6x – 8y + 1 = 0 is:
(a) (0, 0)
(b) (3, – 4)
(c) (– 3, 4) *
(d) (6, – 8)


how can i get this answer ?
Answered by Steve
regroup to form perfect squares:

x^2 + 6x + y^2 - 8y + 1 = 0
x^2+6x+9 + y^2-8y+16 + 1 = 9+16
(x+3)^2 + (Y-4)^2 = 24

center is (-3,4)
Answered by hanan
from where did you bring 9+16??