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

The radius 2x^2+2y^2-17x=0 is r= 17/4

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

radius

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

And the radius is:

r = \frac{17}{4}

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5 x^{2} - 20 x + 5 y^{2} + 30 y - 11 = 0