What is 2x^2+2y^2-8x+28y+86=0 ?

The solution to 2x^2+2y^2-8x+28y+86=0 is Circle with (a,b)=(2,-7),r=sqrt(10)

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2x^2+2y^2-8x+28y+86=0

2 x^{2} + 2 y^{2} - 8 x + 28 y + 86 = 0

(a, b) = (2, - 7), r = 10

Solution steps

2 x^{2} + 2 y^{2} - 8 x + 28 y + 86 = 0

Rewrite

2 x^{2} + 2 y^{2} - 8 x + 28 y + 86 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 7))^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (2, - 7), r = 10

(x + 12)^{2} + (Y (x) - 4)^{2} - 30 = 0

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

x^{2} + y^{2} - x + 6 y + 9 = 0

x^{2} + y^{2} + 24 x + 10 y + 160 = 0

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

What is 2x^2+2y^2-8x+28y+86=0 ?
The solution to 2x^2+2y^2-8x+28y+86=0 is Circle with (a,b)=(2,-7),r=sqrt(10)