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

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

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

center

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

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

Solution steps

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

Rewrite

4 x^{2} + 12 x + 4 y^{2} + 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 - 3)^{2} + (y - 5)^{2} = 4

x^{2} + 20 x + y^{2} + 16 y = - 20

100 x^{2} + 100 y^{2} - 100 x + 400 y + 409 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

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

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