What is the center 4x^2+4y^2-8x-4y=11 ?

The center 4x^2+4y^2-8x-4y=11 is (1, 1/2)

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center 4x^2+4y^2-8x-4y=11

center

4 x^{2} + 4 y^{2} - 8 x - 4 y = 11

(1, \frac{1}{2})

Solution steps

4 x^{2} + 4 y^{2} - 8 x - 4 y = 11

Rewrite

4 x^{2} + 4 y^{2} - 8 x - 4 y = 11

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, \frac{1}{2}), r = 2

And the center is:

(1, \frac{1}{2})

(x + 5)^{2} + (y + 4)^{2} = 9

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

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

(x - 5)^{2} + (y - 2)^{2} = 25

9 x^{2} + 9 y^{2} - 6 x + 24 y = 19

What is the center 4x^2+4y^2-8x-4y=11 ?
The center 4x^2+4y^2-8x-4y=11 is (1, 1/2)