What is the center x^2+y^2-6x+7y-16=0 ?

The center x^2+y^2-6x+7y-16=0 is (3,-7/2)

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center x^2+y^2-6x+7y-16=0

center

x^{2} + y^{2} - 6 x + 7 y - 16 = 0

(3, - \frac{7}{2})

Solution steps

x^{2} + y^{2} - 6 x + 7 y - 16 = 0

Rewrite

x^{2} + y^{2} - 6 x + 7 y - 16 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- \frac{7}{2}))^{2} = (\frac{149}{2})^{2}

Therefore the circle properties are:

(a, b) = (3, - \frac{7}{2}), r = \frac{149}{2}

And the center is:

(3, - \frac{7}{2})

(x - 5)^{2} + (y - 1)^{2} = 49

x^{2} + (y - 1)^{2} = 36

x^{2} + 4 x + 10 y + y^{2} - 2 = 0

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

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

What is the center x^2+y^2-6x+7y-16=0 ?
The center x^2+y^2-6x+7y-16=0 is (3,-7/2)