What is the center x^2+y^2-6x-8y+9=0 ?

The center x^2+y^2-6x-8y+9=0 is (3,4)

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center x^2+y^2-6x-8y+9=0

center

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

(3, 4)

Solution steps

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

Rewrite

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(3, 4)

(x + 2)^{2} + (y - 4)^{2} = 16

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

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