What is the center x^2+y^2-20x+64y-371=0 ?

The center x^2+y^2-20x+64y-371=0 is (10,-32)

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center x^2+y^2-20x+64y-371=0

center

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

(10, - 32)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 10)^{2} + (y - (- 32))^{2} = (1495)^{2}

Therefore the circle properties are:

(a, b) = (10, - 32), r = 1495

And the center is:

(10, - 32)

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

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

2 x^{2} + 2 y^{2} - 4 x + 12 y - 12 = 0

(x + 2)^{2} + y^{2} = 16

x^{2} + y^{2} - 12 x - 12 y = 4

What is the center x^2+y^2-20x+64y-371=0 ?
The center x^2+y^2-20x+64y-371=0 is (10,-32)