What is the center 9x^2-9y^2-18x+54=0 ?

The center 9x^2-9y^2-18x+54=0 is (1,0)

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center 9x^2-9y^2-18x+54=0

center

9 x^{2} - 9 y^{2} - 18 x + 54 = 0

(1, 0)

Solution steps

9 x^{2} - 9 y^{2} - 18 x + 54 = 0

Rewrite

9 x^{2} - 9 y^{2} - 18 x + 54 = 0

in the form of a standard hyperbola equation

\frac{( y - 0 ) ^{2}}{( 5 ) ^{2}} - \frac{( x - 1 ) ^{2}}{( 5 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (1, 0), a = 5, b = 5

And the center is:

(1, 0)

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 5 ) ^{2}}{25} = 1

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

\frac{( x - 3 ) ^{2}}{a ^{2}} - \frac{( y - 4 ) ^{2}}{b ^{2}} = 1

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

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1