What is the center (x+5)^2+(y+2)^2=29 ?

The center (x+5)^2+(y+2)^2=29 is (-5,-2)

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center (x+5)^2+(y+2)^2=29

center

(x + 5)^{2} + (y + 2)^{2} = 29

(- 5, - 2)

Solution steps

(x + 5)^{2} + (y + 2)^{2} = 29

Rewrite

(x + 5)^{2} + (y + 2)^{2} = 29

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 2))^{2} = (29)^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 2), r = 29

And the center is:

(- 5, - 2)

x^{2} + y^{2} - 16 x + 12 y + 64 = 0

x^{2} + 2 x - 4 y + y^{2} = 76

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

x^{2} + y^{2} + 6 x - 8 y + 16 = 0

9 x^{2} + 9 y^{2} = 729

What is the center (x+5)^2+(y+2)^2=29 ?
The center (x+5)^2+(y+2)^2=29 is (-5,-2)