What is 2x^2+2y^2+28y-8y+4=0 ?

The solution to 2x^2+2y^2+28y-8y+4=0 is Circle with (a,b)=(0,-5),r=sqrt(23)

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2x^2+2y^2+28y-8y+4=0

2 x^{2} + 2 y^{2} + 28 y - 8 y + 4 = 0

(a, b) = (0, - 5), r = 23

Solution steps

2 x^{2} + 2 y^{2} + 28 y - 8 y + 4 = 0

Rewrite

2 x^{2} + 2 y^{2} + 28 y - 8 y + 4 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 5))^{2} = (23)^{2}

Therefore the circle properties are:

(a, b) = (0, - 5), r = 23

x^{2} + (y - 9)^{2} = 1

x^{2} + y^{2} + 5 x + 4 y - 34 = 0

(x + 7)^{2} + y^{2} = 36

(x + 6)^{2} + (y - 4)^{2} = 73

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

What is 2x^2+2y^2+28y-8y+4=0 ?
The solution to 2x^2+2y^2+28y-8y+4=0 is Circle with (a,b)=(0,-5),r=sqrt(23)