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

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

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

center

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

(- \frac{1}{2}, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - 2)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, 2), r = 3

And the center is:

(- \frac{1}{2}, 2)

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

(x - 6)^{2} + (y + 5)^{2} = 16

x^{2} + y^{2} + 20 x - 18 y + 145 = 0

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

x^{2} + y^{2} - 3 x - 4 y + 4 = 0

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