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

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

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

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

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

Solution steps

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

Rewrite

2 x^{2} + 8 x + 2 y^{2} + 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 - 5)^{2} + (y - 7)^{2} = 16

(x + 5)^{2} + (y - 3)^{2} = 1

x^{2} + y^{2} = x + 11 y

(x - 4)^{2} + (y + 5)^{2} = 25

(x - 4)^{2} + (y + 5)^{2} = 16

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