What is the center x^2+y^2-8x-14y-16=0 ?

The center x^2+y^2-8x-14y-16=0 is (4,7)

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center x^2+y^2-8x-14y-16=0

center

x^{2} + y^{2} - 8 x - 14 y - 16 = 0

(4, 7)

Solution steps

x^{2} + y^{2} - 8 x - 14 y - 16 = 0

Rewrite

x^{2} + y^{2} - 8 x - 14 y - 16 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 7)^{2} = 9^{2}

Therefore the circle properties are:

(a, b) = (4, 7), r = 9

And the center is:

(4, 7)

(x - 3)^{2} + (y + 2)^{2} = 29

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

(x - 1)^{2} + (y + 4)^{2} = 25

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

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

What is the center x^2+y^2-8x-14y-16=0 ?
The center x^2+y^2-8x-14y-16=0 is (4,7)