What is the center x^2+y^2-14x+6y+49=0 ?

The center x^2+y^2-14x+6y+49=0 is (7,-3)

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center x^2+y^2-14x+6y+49=0

center

x^{2} + y^{2} - 14 x + 6 y + 49 = 0

(7, - 3)

Solution steps

x^{2} + y^{2} - 14 x + 6 y + 49 = 0

Rewrite

x^{2} + y^{2} - 14 x + 6 y + 49 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 3))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (7, - 3), r = 3

And the center is:

(7, - 3)

X^{2} + Y^{2} = 4

(x + 4)^{2} + (y - 1)^{2} = 18

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

3 x^{2} + 3 y^{2} = 75

x^{2} + y^{2} - 5 x + 4 y - 46 = 0

What is the center x^2+y^2-14x+6y+49=0 ?
The center x^2+y^2-14x+6y+49=0 is (7,-3)