What is the center x^2+y^2+12x+8y+3=0 ?

The center x^2+y^2+12x+8y+3=0 is (-6,-4)

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center x^2+y^2+12x+8y+3=0

center

x^{2} + y^{2} + 12 x + 8 y + 3 = 0

(- 6, - 4)

Solution steps

x^{2} + y^{2} + 12 x + 8 y + 3 = 0

Rewrite

x^{2} + y^{2} + 12 x + 8 y + 3 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - (- 4))^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (- 6, - 4), r = 7

And the center is:

(- 6, - 4)

(x + 4)^{2} + (y - 5)^{2} = 25

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

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

(x - 6)^{2} + (y + 5)^{2} = 16

x^{2} + y^{2} + 20 x - 18 y + 145 = 0

What is the center x^2+y^2+12x+8y+3=0 ?
The center x^2+y^2+12x+8y+3=0 is (-6,-4)