What is the center (x-13)^2+(y+12)^2=64 ?

The center (x-13)^2+(y+12)^2=64 is (13,-12)

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center (x-13)^2+(y+12)^2=64

center

(x - 13)^{2} + (y + 12)^{2} = 64

(13, - 12)

Solution steps

(x - 13)^{2} + (y + 12)^{2} = 64

Rewrite

(x - 13)^{2} + (y + 12)^{2} = 64

in the form of the standard circle equation

(x - 13)^{2} + (y - (- 12))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (13, - 12), r = 8

And the center is:

(13, - 12)

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

9 x^{2} - 42 x + 9 y^{2} + 84 y + 65 = 0

x^{2} + y^{2} - 8 y + 7 = 0

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

What is the center (x-13)^2+(y+12)^2=64 ?
The center (x-13)^2+(y+12)^2=64 is (13,-12)