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

The center x^2+6x+1+y^2=4 is (-3,0)

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center x^2+6x+1+y^2=4

center

x^{2} + 6 x + 1 + y^{2} = 4

(- 3, 0)

Solution steps

x^{2} + 6 x + 1 + y^{2} = 4

Rewrite

x^{2} + 6 x + 1 + y^{2} = 4

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 0)^{2} = (23)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 0), r = 23

And the center is:

(- 3, 0)

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

(x - 12)^{2} + (y - 7)^{2} = 196

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

(x - 2)^{2} + (y - 5)^{2} = 49

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1