What is 2x^2+2y^2-4x-8y=10 ?

The solution to 2x^2+2y^2-4x-8y=10 is Circle with (a,b)=(1,2),r=sqrt(10)

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2x^2+2y^2-4x-8y=10

2 x^{2} + 2 y^{2} - 4 x - 8 y = 10

(a, b) = (1, 2), r = 10

Solution steps

2 x^{2} + 2 y^{2} - 4 x - 8 y = 10

Rewrite

2 x^{2} + 2 y^{2} - 4 x - 8 y = 10

in the form of the standard circle equation

(x - 1)^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (1, 2), r = 10

3 x^{2} + 3 y^{2} - 1 = 0

x^{2} + y^{2} - 12 x + 10 y + 23 = 0

x^{2} + y^{2} + 10 x = 11 - 2 y

x^{2} + (y - 5)^{2} = 81

2 x^{2} - 8 x + 2 y^{2} + 12 y + 14 = 0

What is 2x^2+2y^2-4x-8y=10 ?
The solution to 2x^2+2y^2-4x-8y=10 is Circle with (a,b)=(1,2),r=sqrt(10)