What is 2x^2+2y^2-28x+12y+114=0 ?

The solution to 2x^2+2y^2-28x+12y+114=0 is Circle with (a,b)=(7,-3),r=1

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2x^2+2y^2-28x+12y+114=0

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

(a, b) = (7, - 3), r = 1

Solution steps

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

Rewrite

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 3))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (7, - 3), r = 1

(x - 4)^{2} + (y + 11)^{2} = 25

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

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

x^{2} + y^{2} + 2 x - 244 = 0

(x - 3)^{2} + y^{2} < 4

What is 2x^2+2y^2-28x+12y+114=0 ?
The solution to 2x^2+2y^2-28x+12y+114=0 is Circle with (a,b)=(7,-3),r=1