What is the center 4x^2-4x+4y^2-63=0 ?

The center 4x^2-4x+4y^2-63=0 is (1/2 ,0)

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center 4x^2-4x+4y^2-63=0

center

4 x^{2} - 4 x + 4 y^{2} - 63 = 0

(\frac{1}{2}, 0)

Solution steps

4 x^{2} - 4 x + 4 y^{2} - 63 = 0

Rewrite

4 x^{2} - 4 x + 4 y^{2} - 63 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 0), r = 4

And the center is:

(\frac{1}{2}, 0)

(x - 3)^{2} + (y + 4)^{2} = 25

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

16 x^{2} + 16 y^{2} + 96 x - 40 y = 315

x^{2} + y^{2} - 2 x + 6 y + 9 = 0

x^{2} + y^{2} - 4 x + 10 y + 13 = 0