What is 2x^2+8x+2y^2-8y-4=0 ?

The solution to 2x^2+8x+2y^2-8y-4=0 is Circle with (a,b)=(-2,2),r=sqrt(10)

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2x^2+8x+2y^2-8y-4=0

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

(a, b) = (- 2, 2), r = 10

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (- 2, 2), r = 10

x^{2} + y^{2} - 8 x - 12 y + 27 = 0

4 = (6 - x)^{2} + (2 - y)^{2}

9 x^{2} + 9 y^{2} = 3

x (x + 5) + (y + 2)^{2} = 0

x^{2} + y^{2} - 2 x + 6 y - 111 = 0

What is 2x^2+8x+2y^2-8y-4=0 ?
The solution to 2x^2+8x+2y^2-8y-4=0 is Circle with (a,b)=(-2,2),r=sqrt(10)