What is the center 100x^2+100y^2-100x+400y+409=0 ?

The center 100x^2+100y^2-100x+400y+409=0 is (1/2 ,-2)

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center 100x^2+100y^2-100x+400y+409=0

center

(\frac{1}{2}, - 2)

Solution steps

100 x^{2} + 100 y^{2} - 100 x + 400 y + 409 = 0

Rewrite

100 x^{2} + 100 y^{2} - 100 x + 400 y + 409 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - 2), r = \frac{2}{5}

And the center is:

(\frac{1}{2}, - 2)

What is the center 100x^2+100y^2-100x+400y+409=0 ?
The center 100x^2+100y^2-100x+400y+409=0 is (1/2 ,-2)