What is the center x^2-14x+y^2+8y+40=0 ?

The center x^2-14x+y^2+8y+40=0 is (7,-4)

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center x^2-14x+y^2+8y+40=0

center

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

(7, - 4)

Solution steps

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

Rewrite

x^{2} - 14 x + y^{2} + 8 y + 40 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 4))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 5

And the center is:

(7, - 4)

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

x^{2} + y^{2} + 6 y + 4 x + 8 = 0

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

x^{2} + y^{2} - 8 x - 2 y - 19 = 0

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

What is the center x^2-14x+y^2+8y+40=0 ?
The center x^2-14x+y^2+8y+40=0 is (7,-4)