What is 2x^2+8x+2y^2+3y=0 ?

The solution to 2x^2+8x+2y^2+3y=0 is Circle with (a,b)=(-2,-3/4),r=(sqrt(73))/4

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2x^2+8x+2y^2+3y=0

2 x^{2} + 8 x + 2 y^{2} + 3 y = 0

(a, b) = (- 2, - \frac{3}{4}), r = \frac{73}{4}

Solution steps

2 x^{2} + 8 x + 2 y^{2} + 3 y = 0

Rewrite

2 x^{2} + 8 x + 2 y^{2} + 3 y = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- \frac{3}{4}))^{2} = (\frac{73}{4})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - \frac{3}{4}), r = \frac{73}{4}

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- y^{(2)} + 14 y = x^{(2)} - 10 x + 25 + 49 - 12.24

What is 2x^2+8x+2y^2+3y=0 ?
The solution to 2x^2+8x+2y^2+3y=0 is Circle with (a,b)=(-2,-3/4),r=(sqrt(73))/4