What is the center (x+9/4)^2= 13/2-(y-8)^2 ?

The center (x+9/4)^2= 13/2-(y-8)^2 is (-9/4 ,8)

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center (x+9/4)^2= 13/2-(y-8)^2

center

(x + \frac{9}{4})^{2} = \frac{13}{2} - (y - 8)^{2}

(- \frac{9}{4}, 8)

Solution steps

(x + \frac{9}{4})^{2} = \frac{13}{2} - (y - 8)^{2}

Rewrite

(x + \frac{9}{4})^{2} = \frac{13}{2} - (y - 8)^{2}

in the form of the standard circle equation

(x - (- \frac{9}{4}))^{2} + (y - 8)^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{9}{4}, 8), r = \frac{13}{2}

And the center is:

(- \frac{9}{4}, 8)

x^{2} + y^{2} + 12 x + 8 y + 3 = 0

(x + 4)^{2} + (y - 5)^{2} = 25

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

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

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

What is the center (x+9/4)^2= 13/2-(y-8)^2 ?
The center (x+9/4)^2= 13/2-(y-8)^2 is (-9/4 ,8)