What is the center (x+2)^2+(y-sqrt(5))^2=7 ?

The center (x+2)^2+(y-sqrt(5))^2=7 is (-2,sqrt(5))

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center (x+2)^2+(y-sqrt(5))^2=7

center

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

(- 2, 5)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 5), r = 7

And the center is:

(- 2, 5)

x (x - 12) + y (y + 2) = 12

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

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

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

4 x^{2} + 40 x + 4 y^{2} = 0

What is the center (x+2)^2+(y-sqrt(5))^2=7 ?
The center (x+2)^2+(y-sqrt(5))^2=7 is (-2,sqrt(5))