What is the center x^2+y^2-10x+4y+20=0 ?

The center x^2+y^2-10x+4y+20=0 is (5,-2)

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center x^2+y^2-10x+4y+20=0

center

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

(5, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (5, - 2), r = 3

And the center is:

(5, - 2)

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

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

x^{2} + y^{2} + 14 x - 16 + 64 = 0

x^{2} + 8 x + y^{2} + 4 y = 4

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

What is the center x^2+y^2-10x+4y+20=0 ?
The center x^2+y^2-10x+4y+20=0 is (5,-2)