What is the center 2x^2+2y^2+6x-8y+12=0 ?

The center 2x^2+2y^2+6x-8y+12=0 is (-3/2 ,2)

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center 2x^2+2y^2+6x-8y+12=0

center

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

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

Solution steps

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

Rewrite

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 2), r = \frac{1}{2}

And the center is:

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

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

(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

What is the center 2x^2+2y^2+6x-8y+12=0 ?
The center 2x^2+2y^2+6x-8y+12=0 is (-3/2 ,2)