What is the center x^2+y^2-6x+4y-3=0 ?

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

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center x^2+y^2-6x+4y-3=0

center

x^{2} + y^{2} - 6 x + 4 y - 3 = 0

(3, - 2)

Solution steps

x^{2} + y^{2} - 6 x + 4 y - 3 = 0

Rewrite

x^{2} + y^{2} - 6 x + 4 y - 3 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(3, - 2)

(x + 12)^{2} + (y - 9)^{2} = 35

x^{2} + y^{2} = 10

(x + 1)^{2} + (y + 3)^{2} = 5

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

(x + 3)^{2} + (y - 1)^{2} = 9