What is the center x^2-7x+y^2+8y-14=0 ?

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

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center x^2-7x+y^2+8y-14=0

center

(\frac{7}{2}, - 4)

Solution steps

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

Rewrite

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

in the form of the standard circle equation

(x - \frac{7}{2})^{2} + (y - (- 4))^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{7}{2}, - 4), r = \frac{13}{2}

And the center is:

(\frac{7}{2}, - 4)

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