What is the center (x+3)^2+y^2=48 ?

The center (x+3)^2+y^2=48 is (-3,0)

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center (x+3)^2+y^2=48

center

(x + 3)^{2} + y^{2} = 48

(- 3, 0)

Solution steps

(x + 3)^{2} + y^{2} = 48

Rewrite

(x + 3)^{2} + y^{2} = 48

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 0)^{2} = (48)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 0), r = 48

And the center is:

(- 3, 0)

x^{2} + y^{2} + 2 x - \frac{27}{2} y + 20 = 0

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

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

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

x^{2} + y^{2} = 289