What is the center x^2+y^2-8x-12y+43=0 ?

The center x^2+y^2-8x-12y+43=0 is (4,6)

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center x^2+y^2-8x-12y+43=0

center

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

(4, 6)

Solution steps

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

Rewrite

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 6), r = 3

And the center is:

(4, 6)

(x - 1)^{2} + (y + 3)^{2} = 25

(x + 14)^{2} + (y - 11)^{2} = 16

x^{2} + y^{2} - 6 x + 10 y + 9 = 0

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

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

What is the center x^2+y^2-8x-12y+43=0 ?
The center x^2+y^2-8x-12y+43=0 is (4,6)