What is the center (x+4)^2+(y+3)^2=25 ?

The center (x+4)^2+(y+3)^2=25 is (-4,-3)

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center (x+4)^2+(y+3)^2=25

center

(x + 4)^{2} + (y + 3)^{2} = 25

(- 4, - 3)

Solution steps

(x + 4)^{2} + (y + 3)^{2} = 25

Rewrite

(x + 4)^{2} + (y + 3)^{2} = 25

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 3))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 3), r = 5

And the center is:

(- 4, - 3)

(x - 1)^{2} + (y - 3)^{2} = 9

5 x^{2} + 3 x + 5 y^{2} - 2 y - 9 = 0

(x - 3)^{2} + (y - 1)^{2} = 16

x^{2} + y^{2} + 2 x + 4 y = 11

x^{2} + y^{2} + 2 y - 16 = 0

What is the center (x+4)^2+(y+3)^2=25 ?
The center (x+4)^2+(y+3)^2=25 is (-4,-3)