What is the center 4x^2+4y^2+12x+16y-11=0 ?

The center 4x^2+4y^2+12x+16y-11=0 is (-3/2 ,-2)

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center 4x^2+4y^2+12x+16y-11=0

center

4 x^{2} + 4 y^{2} + 12 x + 16 y - 11 = 0

(- \frac{3}{2}, - 2)

Solution steps

4 x^{2} + 4 y^{2} + 12 x + 16 y - 11 = 0

Rewrite

4 x^{2} + 4 y^{2} + 12 x + 16 y - 11 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, - 2), r = 3

And the center is:

(- \frac{3}{2}, - 2)

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

x^{2} + y^{2} + 2 x = 3 y

x^{2} + 8 x + y^{2} + 2 y + 8 = 0

x^{2} + 6 x + y^{2} - 3 y = - \frac{1}{4}

x^{2} + y^{2} - 3 x + 4 y = 0

What is the center 4x^2+4y^2+12x+16y-11=0 ?
The center 4x^2+4y^2+12x+16y-11=0 is (-3/2 ,-2)