What is 2x^2+2y^2-3x=0 ?

The solution to 2x^2+2y^2-3x=0 is Circle with (a,b)=(3/4 ,0),r= 3/4

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2x^2+2y^2-3x=0

2 x^{2} + 2 y^{2} - 3 x = 0

(a, b) = (\frac{3}{4}, 0), r = \frac{3}{4}

Solution steps

2 x^{2} + 2 y^{2} - 3 x = 0

Rewrite

2 x^{2} + 2 y^{2} - 3 x = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - 0)^{2} = (\frac{3}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, 0), r = \frac{3}{4}

x^{2} + y^{2} - 14 y + 45 = 0

x^{2} + y^{2} - 8 x + 12 y = - 43

x^{2} + z^{2} = 9

(x + 5)^{2} + (y - 1)^{2} = 9

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

What is 2x^2+2y^2-3x=0 ?
The solution to 2x^2+2y^2-3x=0 is Circle with (a,b)=(3/4 ,0),r= 3/4