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

(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)

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)