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 ?
To find the center of a circle given its equation, we need to rewrite the equation in the standard form of a circle, which is:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the center of the circle, and r is the radius.
Now let's convert the given equation to the standard form by completing the square:
x^2 + y^2 + 6x - 8y + 1 = 0
Rearranging the terms:
x^2 + 6x + y^2 - 8y = -1
To complete the square for the x terms, we add (6/2)^2 = 9 to both sides, and for the y terms, we add (-8/2)^2 = 16 to both sides:
x^2 + 6x + 9 + y^2 - 8y + 16 = -1 + 9 + 16
Simplifying:
(x + 3)^2 + (y - 4)^2 = 24
Comparing this form to the standard form, we can see that the center (h, k) is (-3, 4). Therefore, the correct answer is option (c) (-3, 4).
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)