What is the center+x^2+y^2-14x+2y+41=0 ?

The center+x^2+y^2-14x+2y+41=0 is (7,-1)

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center+x^2+y^2-14x+2y+41=0

center

+ x^{2} + y^{2} - 14 x + 2 y + 41 = 0

(7, - 1)

Solution steps

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

Rewrite

+ x^{2} + y^{2} - 14 x + 2 y + 41 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 1))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (7, - 1), r = 3

And the center is:

(7, - 1)

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

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

(x - 3)^{2} + (y + 5)^{2} = 25

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

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

What is the center+x^2+y^2-14x+2y+41=0 ?
The center+x^2+y^2-14x+2y+41=0 is (7,-1)