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

The solution to 5x^2+9x+5y^2-3y-5=0 is Circle with (a,b)=(-9/10 , 3/10),r=sqrt(19/10)

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

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

(a, b) = (- \frac{9}{10}, \frac{3}{10}), r = \frac{19}{10}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{9}{10}))^{2} + (y - \frac{3}{10})^{2} = (\frac{19}{10})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{9}{10}, \frac{3}{10}), r = \frac{19}{10}

(x - 2)^{2} + (y + 4)^{2} = 9

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

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

x^{2} + y^{2} + 14 x - 12 y + 4 = 0

3 x^{2} + 3 y^{2} - 6 y - 1 = 0

What is 5x^2+9x+5y^2-3y-5=0 ?
The solution to 5x^2+9x+5y^2-3y-5=0 is Circle with (a,b)=(-9/10 , 3/10),r=sqrt(19/10)