What is the center x^2+y^2+10x-8y+25=0 ?

The center x^2+y^2+10x-8y+25=0 is (-5,4)

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center x^2+y^2+10x-8y+25=0

center

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

(- 5, 4)

Solution steps

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

Rewrite

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 5, 4), r = 4

And the center is:

(- 5, 4)

4 x^{2} + 4 y^{2} = 144

(x + 1)^{2} + y^{2} = 246

(x - 4)^{2} + (y + 1^{2})^{2} = 9

(x + \frac{9}{4})^{2} = \frac{13}{2} - (y - 8)^{2}

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

What is the center x^2+y^2+10x-8y+25=0 ?
The center x^2+y^2+10x-8y+25=0 is (-5,4)