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

The solution to 2x^2+2y^2+5x+3y+2=0 is Circle with (a,b)=(-5/4 ,-3/4),r= 3/(2sqrt(2))

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

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

(a, b) = (- \frac{5}{4}, - \frac{3}{4}), r = \frac{3}{2 2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{5}{4}))^{2} + (y - (- \frac{3}{4}))^{2} = (\frac{3}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{4}, - \frac{3}{4}), r = \frac{3}{2 2}

(y + 4)^{2} + (x + 3)^{2} = 25

16 x^{2} + 96 x + 16 y^{2} - 40 y = 315

x^{2} - 4 x + y^{2} + 8 y = - 4

x^{2} + y^{2} + 4 y - 1 = 0

x^{2} + y^{2} + 4 x - 8 y + 9 = 0

What is 2x^2+2y^2+5x+3y+2=0 ?
The solution to 2x^2+2y^2+5x+3y+2=0 is Circle with (a,b)=(-5/4 ,-3/4),r= 3/(2sqrt(2))