What is the center x^2+6x+y^2-3y=-1/4 ?

The center x^2+6x+y^2-3y=-1/4 is (-3, 3/2)

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center x^2+6x+y^2-3y=-1/4

center

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{3}{2})^{2} = (11)^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{3}{2}), r = 11

And the center is:

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

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

x^{2} + 4 x + 10 y + y^{2} + 3 = 0

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

x^{2} + y^{2} - 6 x + 5 = 0

(x - 1)^{2} + (y + 4)^{2} = 36

What is the center x^2+6x+y^2-3y=-1/4 ?
The center x^2+6x+y^2-3y=-1/4 is (-3, 3/2)