What is the center 2x^2+2y^2-5x+4y-1=0 ?

The center 2x^2+2y^2-5x+4y-1=0 is (5/4 ,-1)

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center 2x^2+2y^2-5x+4y-1=0

center

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

(\frac{5}{4}, - 1)

Solution steps

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

Rewrite

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

in the form of the standard circle equation

(x - \frac{5}{4})^{2} + (y - (- 1))^{2} = (\frac{7}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{4}, - 1), r = \frac{7}{4}

And the center is:

(\frac{5}{4}, - 1)

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

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

x^{2} + y^{2} = 21

x^{2} + y^{2} - 20 x + 36 y + 353 = 0

(x + 1)^{2} + (y - 5)^{2} = 4

What is the center 2x^2+2y^2-5x+4y-1=0 ?
The center 2x^2+2y^2-5x+4y-1=0 is (5/4 ,-1)