What is the center 4x^2+4y^2+40y+75=0 ?

The center 4x^2+4y^2+40y+75=0 is (0,-5)

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center 4x^2+4y^2+40y+75=0

center

4 x^{2} + 4 y^{2} + 40 y + 75 = 0

(0, - 5)

Solution steps

4 x^{2} + 4 y^{2} + 40 y + 75 = 0

Rewrite

4 x^{2} + 4 y^{2} + 40 y + 75 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(0, - 5)

(x + \frac{8}{3})^{2} + y^{2} = 1

(x - 3)^{2} + (y + 4)^{2} = \frac{16}{9}

x^{2} + y^{2} + 4 y - 5 = 0

(x + 2)^{2} + (y - 3)^{2} = 15

(x + 5)^{2} + (y + 2)^{2} = 29