What is the center 4x^2+4y^2=25 ?

The center 4x^2+4y^2=25 is (0,0)

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center 4x^2+4y^2=25

center

4 x^{2} + 4 y^{2} = 25

(0, 0)

Solution steps

4 x^{2} + 4 y^{2} = 25

Rewrite

4 x^{2} + 4 y^{2} = 25

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = \frac{5}{2}

And the center is:

(0, 0)

x^{2} + y^{2} = 64

x^{2} + y^{2} + 4 x - 6 y + 12 = 0

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

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

2 x^{2} + 2 y^{2} - 12 x + 8 y - 24 = 0