What is the center x^2+y^2+5x+4y-34=0 ?

The center x^2+y^2+5x+4y-34=0 is (-5/2 ,-2)

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center x^2+y^2+5x+4y-34=0

center

x^{2} + y^{2} + 5 x + 4 y - 34 = 0

(- \frac{5}{2}, - 2)

Solution steps

x^{2} + y^{2} + 5 x + 4 y - 34 = 0

Rewrite

x^{2} + y^{2} + 5 x + 4 y - 34 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - (- 2))^{2} = (\frac{177}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, - 2), r = \frac{177}{2}

And the center is:

(- \frac{5}{2}, - 2)

(x - 4)^{2} + (y - 2)^{2} = 49

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

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

(x + 6)^{2} + (y - 1)^{2} = 16

(x + 6)^{2} + (y - 7)^{2} = 81

What is the center x^2+y^2+5x+4y-34=0 ?
The center x^2+y^2+5x+4y-34=0 is (-5/2 ,-2)