What is the center 4x^2+4y^2-16x+40y+76=0 ?

The center 4x^2+4y^2-16x+40y+76=0 is (2,-5)

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center 4x^2+4y^2-16x+40y+76=0

center

4 x^{2} + 4 y^{2} - 16 x + 40 y + 76 = 0

(2, - 5)

Solution steps

4 x^{2} + 4 y^{2} - 16 x + 40 y + 76 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 x + 40 y + 76 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 5))^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (2, - 5), r = 10

And the center is:

(2, - 5)

x^{2} + y^{2} \leq 9

x^{2} + 10 x + y^{2} - 16 - 11 = 0

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

4 x^{2} + 4 y^{2} = 144

(x + 1)^{2} + y^{2} = 246

What is the center 4x^2+4y^2-16x+40y+76=0 ?
The center 4x^2+4y^2-16x+40y+76=0 is (2,-5)