What is the center (x+3)^2-49=-y^2 ?

The center (x+3)^2-49=-y^2 is (-3,0)

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center (x+3)^2-49=-y^2

center

(x + 3)^{2} - 49 = - y^{2}

(- 3, 0)

Solution steps

(x + 3)^{2} - 49 = - y^{2}

Rewrite

(x + 3)^{2} - 49 = - y^{2}

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 0)^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (- 3, 0), r = 7

And the center is:

(- 3, 0)

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

(x + \frac{5}{2})^{2} + (y - 14)^{2} = 20

x^{2} + y^{2} = 100

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

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