What is the center (x^2)/9+(y^2)/9 =1 ?

The center (x^2)/9+(y^2)/9 =1 is (0,0)

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center (x^2)/9+(y^2)/9 =1

center

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

Rewrite

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(0, 0)

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

(x + 3)^{2} + (y - 4)^{2} = 16

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

2 x^{2} + 2 y^{2} = 8

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{9}{4}