What is 2x^2-8x+8+2y^2+4x+2=25 ?

The solution to 2x^2-8x+8+2y^2+4x+2=25 is Circle with (a,b)=(1,0),r=sqrt(17/2)

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2x^2-8x+8+2y^2+4x+2=25

2 x^{2} - 8 x + 8 + 2 y^{2} + 4 x + 2 = 25

(a, b) = (1, 0), r = \frac{17}{2}

Solution steps

2 x^{2} - 8 x + 8 + 2 y^{2} + 4 x + 2 = 25

Rewrite

2 x^{2} - 8 x + 8 + 2 y^{2} + 4 x + 2 = 25

in the form of the standard circle equation

(x - 1)^{2} + (y - 0)^{2} = (\frac{17}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, 0), r = \frac{17}{2}

100 = x^{2} + y^{2} + 4 x + 60

(x + 3)^{2} + (y + 3)^{2} = 25

(x - 8)^{2} + (y - 3)^{2} = 9

(x - 4.5)^{2} + (y - 1.5)^{2} = 5.52

x^{2} + y^{2} + 2 x + 1 < 1

What is 2x^2-8x+8+2y^2+4x+2=25 ?
The solution to 2x^2-8x+8+2y^2+4x+2=25 is Circle with (a,b)=(1,0),r=sqrt(17/2)