What is the center x^2+y^2-4x-5=0 ?

The center x^2+y^2-4x-5=0 is (2,0)

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center x^2+y^2-4x-5=0

center

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

(2, 0)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = 3

And the center is:

(2, 0)

(x - 3)^{2} + (y + 5)^{2} = 49

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

x^{2} + y^{2} - 10 x + 4 y + 20 = 0

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}