What is the center x^2+y^2-5x+4y-46=0 ?

The center x^2+y^2-5x+4y-46=0 is (5/2 ,-2)

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center x^2+y^2-5x+4y-46=0

center

x^{2} + y^{2} - 5 x + 4 y - 46 = 0

(\frac{5}{2}, - 2)

Solution steps

x^{2} + y^{2} - 5 x + 4 y - 46 = 0

Rewrite

x^{2} + y^{2} - 5 x + 4 y - 46 = 0

in the form of the standard circle equation

(x - \frac{5}{2})^{2} + (y - (- 2))^{2} = (\frac{15}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, - 2), r = \frac{15}{2}

And the center is:

(\frac{5}{2}, - 2)

3 (x + 1)^{2} + 3 (y - 1)^{2} = 6

x^{2} + y^{2} - 8 x + 12 = 0

x^{2} + y^{2} - 6 x - 8 y - 11 = 0

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

x^{2} + y^{2} + 8 x - 14 y + 1 = 0

What is the center x^2+y^2-5x+4y-46=0 ?
The center x^2+y^2-5x+4y-46=0 is (5/2 ,-2)