What is 2x^2+2y^2+4x-3y=-2 ?

The solution to 2x^2+2y^2+4x-3y=-2 is Circle with (a,b)=(-1, 3/4),r= 3/4

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2x^2+2y^2+4x-3y=-2

2 x^{2} + 2 y^{2} + 4 x - 3 y = - 2

(a, b) = (- 1, \frac{3}{4}), r = \frac{3}{4}

Solution steps

2 x^{2} + 2 y^{2} + 4 x - 3 y = - 2

Rewrite

2 x^{2} + 2 y^{2} + 4 x - 3 y = - 2

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{3}{4})^{2} = (\frac{3}{4})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{3}{4}), r = \frac{3}{4}

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

3 y^{2} + 3 x^{2} - 27 = 0

9 x^{2} - 36 x + 9 y^{2} + 6 y = - 28

x^{2} + y^{2} - 2 x - 16 y + 29 = 0

(x - 3)^{2} + (y - 8)^{2} = 25

What is 2x^2+2y^2+4x-3y=-2 ?
The solution to 2x^2+2y^2+4x-3y=-2 is Circle with (a,b)=(-1, 3/4),r= 3/4