What is the center (x+3)^2+(y-4)^2=7 ?

The center (x+3)^2+(y-4)^2=7 is (-3,4)

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center (x+3)^2+(y-4)^2=7

center

(x + 3)^{2} + (y - 4)^{2} = 7

(- 3, 4)

Solution steps

(x + 3)^{2} + (y - 4)^{2} = 7

Rewrite

(x + 3)^{2} + (y - 4)^{2} = 7

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 4)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 4), r = 7

And the center is:

(- 3, 4)

x^{2} + y^{2} - 10 x + 6 y - 2 = 0

x^{2} + y^{2} = 3 x + 4 y

(x - 4)^{2} + (y - 5)^{2} = 36

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

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