What is the center 2x^2+2y^2-12x+4y-30=0 ?

The center 2x^2+2y^2-12x+4y-30=0 is (3,-1)

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center 2x^2+2y^2-12x+4y-30=0

center

2 x^{2} + 2 y^{2} - 12 x + 4 y - 30 = 0

(3, - 1)

Solution steps

2 x^{2} + 2 y^{2} - 12 x + 4 y - 30 = 0

Rewrite

2 x^{2} + 2 y^{2} - 12 x + 4 y - 30 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 1))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = 5

And the center is:

(3, - 1)

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

x^{2} - 4 x + y^{2} - 8 y = 20

(x - 3)^{2} + (y + 2)^{2} = \frac{16}{9}

(x - 2)^{2} + y^{2} = 25

3 x^{2} - 15 x = - 3 y^{2}

What is the center 2x^2+2y^2-12x+4y-30=0 ?
The center 2x^2+2y^2-12x+4y-30=0 is (3,-1)