What is the center x^2+y^2+6x+8y+24=0 ?

The center x^2+y^2+6x+8y+24=0 is (-3,-4)

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center x^2+y^2+6x+8y+24=0

center

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

(- 3, - 4)

Solution steps

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

Rewrite

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - (- 4))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 3, - 4), r = 1

And the center is:

(- 3, - 4)

x^{2} + 8 x + y^{2} - 8 y = - 16

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

(x - 3)^{2} + (y - 5)^{2} = 121

(x + 4)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} - 4 x + 6 y - 36 = 0

What is the center x^2+y^2+6x+8y+24=0 ?
The center x^2+y^2+6x+8y+24=0 is (-3,-4)