What is the center 2x^2+2y^2-12x+8y-24=0 ?

The center 2x^2+2y^2-12x+8y-24=0 is (3,-2)

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center 2x^2+2y^2-12x+8y-24=0

center

2 x^{2} + 2 y^{2} - 12 x + 8 y - 24 = 0

(3, - 2)

Solution steps

2 x^{2} + 2 y^{2} - 12 x + 8 y - 24 = 0

Rewrite

2 x^{2} + 2 y^{2} - 12 x + 8 y - 24 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 2), r = 5

And the center is:

(3, - 2)

x^{2} + y^{2} - 6 x + 8 y = 144

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

x^{2} + y^{2} + 6 x - 4 y - 23 = 0

5 x^{2} + 5 y^{2} = 125

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

What is the center 2x^2+2y^2-12x+8y-24=0 ?
The center 2x^2+2y^2-12x+8y-24=0 is (3,-2)