What is 2x^2+2y^2-8x+8y+12=0 ?

The solution to 2x^2+2y^2-8x+8y+12=0 is Circle with (a,b)=(2,-2),r=sqrt(2)

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2x^2+2y^2-8x+8y+12=0

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

y^{2} = 100 - x^{2}

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

y^{2} + x^{2} + 17 - 8 x + 4 y = 0

(x - 4)^{2} + (y - 4)^{2} = 27

What is 2x^2+2y^2-8x+8y+12=0 ?
The solution to 2x^2+2y^2-8x+8y+12=0 is Circle with (a,b)=(2,-2),r=sqrt(2)