What is 2x^2+2y^2-17x=0 ?

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

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2x^2+2y^2-17x=0

2 x^{2} + 2 y^{2} - 17 x = 0

(a, b) = (\frac{17}{4}, 0), r = \frac{17}{4}

Solution steps

2 x^{2} + 2 y^{2} - 17 x = 0

Rewrite

2 x^{2} + 2 y^{2} - 17 x = 0

in the form of the standard circle equation

(x - \frac{17}{4})^{2} + (y - 0)^{2} = (\frac{17}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{17}{4}, 0), r = \frac{17}{4}

x^{2} + y^{2} + 8 x + 6 y - 11 = 0

x^{2} + y^{2} = 28

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

4 x^{2} + 4 y^{2} + 12 x - 4 y - 90 = 0

x^{2} + y^{2} + 6 x - 2 y - 26 = 0

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