What is the center x^2+y^2-8x+4y+11=0 ?

The center x^2+y^2-8x+4y+11=0 is (4,-2)

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center x^2+y^2-8x+4y+11=0

center

x^{2} + y^{2} - 8 x + 4 y + 11 = 0

(4, - 2)

Solution steps

x^{2} + y^{2} - 8 x + 4 y + 11 = 0

Rewrite

x^{2} + y^{2} - 8 x + 4 y + 11 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (4, - 2), r = 3

And the center is:

(4, - 2)

x^{2} + y^{2} - 6 y = 16

x^{2} + y^{2} - 20 x - 20 y + 100 = 0

X^{2} + Y^{2} + 2 X - 6 Y = 0

4 x^{2} + 4 y^{2} = 64

(x - 2)^{2} + (y + 5)^{2} = 9