What is the center 2x^2+2y^2-16x+4y-16=0 ?

The center 2x^2+2y^2-16x+4y-16=0 is (4,-1)

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center 2x^2+2y^2-16x+4y-16=0

center

2 x^{2} + 2 y^{2} - 16 x + 4 y - 16 = 0

(4, - 1)

Solution steps

2 x^{2} + 2 y^{2} - 16 x + 4 y - 16 = 0

Rewrite

2 x^{2} + 2 y^{2} - 16 x + 4 y - 16 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 1))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (4, - 1), r = 5

And the center is:

(4, - 1)

(x - 1)^{2} + (y + 3)^{2} = 16

9 x^{2} + 12 x + 9 y^{2} - 18 y - 23 = 0

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

(x - 1)^{2} + (y + 3)^{2} = 25

(x + 14)^{2} + (y - 11)^{2} = 16

What is the center 2x^2+2y^2-16x+4y-16=0 ?
The center 2x^2+2y^2-16x+4y-16=0 is (4,-1)