What is 1(x^2+y^2)=x ?

The solution to 1(x^2+y^2)=x is Circle with (a,b)=(1/2 ,0),r= 1/2

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1(x^2+y^2)=x

1 (x^{2} + y^{2}) = x

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

Solution steps

1 \cdot (x^{2} + y^{2}) = x

Rewrite

1 (x^{2} + y^{2}) = x

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 2 y - 48 = 0

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

2 x^{2} + 2 y^{2} + 8 x - 12 y - 26 = 0

x^{2} + y^{2} + 12 x - 14 y + 36 = 0

x^{2} + y^{2} - 2 x - 6 y + \frac{1}{4} = 0

What is 1(x^2+y^2)=x ?
The solution to 1(x^2+y^2)=x is Circle with (a,b)=(1/2 ,0),r= 1/2