What is the center 8x^2+8y^2-20x+64y-371=0 ?

The center 8x^2+8y^2-20x+64y-371=0 is (5/4 ,-4)

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center 8x^2+8y^2-20x+64y-371=0

center

(\frac{5}{4}, - 4)

Solution steps

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

Rewrite

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

in the form of the standard circle equation

(x - \frac{5}{4})^{2} + (y - (- 4))^{2} = (\frac{1023}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{4}, - 4), r = \frac{1023}{4}

And the center is:

(\frac{5}{4}, - 4)

What is the center 8x^2+8y^2-20x+64y-371=0 ?
The center 8x^2+8y^2-20x+64y-371=0 is (5/4 ,-4)