What is 2x^2+2y^2+8x+7=0 ?

The solution to 2x^2+2y^2+8x+7=0 is Circle with (a,b)=(-2,0),r=sqrt(1/2)

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2x^2+2y^2+8x+7=0

2 x^{2} + 2 y^{2} + 8 x + 7 = 0

(a, b) = (- 2, 0), r = \frac{1}{2}

Solution steps

2 x^{2} + 2 y^{2} + 8 x + 7 = 0

Rewrite

2 x^{2} + 2 y^{2} + 8 x + 7 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 0)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, 0), r = \frac{1}{2}

x^{2} + z^{2} = 1

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

x^{2} + y^{2} - 12 x + 20 y + 72 = 0

x^{2} + y^{2} - 8 x - 10 y - 12 = 0

x^{2} + y^{2} - 10 x + 2 y = 0

What is 2x^2+2y^2+8x+7=0 ?
The solution to 2x^2+2y^2+8x+7=0 is Circle with (a,b)=(-2,0),r=sqrt(1/2)