What is the center 5x^2+3x+5y^2-2y-9=0 ?

The center 5x^2+3x+5y^2-2y-9=0 is (-3/10 , 1/5)

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center 5x^2+3x+5y^2-2y-9=0

center

(- \frac{3}{10}, \frac{1}{5})

Solution steps

5 x^{2} + 3 x + 5 y^{2} - 2 y - 9 = 0

Rewrite

5 x^{2} + 3 x + 5 y^{2} - 2 y - 9 = 0

in the form of the standard circle equation

(x - (- \frac{3}{10}))^{2} + (y - \frac{1}{5})^{2} = (\frac{193}{10})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{10}, \frac{1}{5}), r = \frac{193}{10}

And the center is:

(- \frac{3}{10}, \frac{1}{5})

What is the center 5x^2+3x+5y^2-2y-9=0 ?
The center 5x^2+3x+5y^2-2y-9=0 is (-3/10 , 1/5)