Answer
$(x-\displaystyle \frac{1}{2})^{2}+y^{2}=\frac{1}{4}$
$x^{2}+y^{2}-x=0$
Work Step by Step
The standard form of an equation of a circle with radius $r$ and center $(h,k)$ is
$(x-h)^{2}+(y-k)^{2}=r^{2}$
When its graph is a circle, the equation
$x^{2}+y^{2}+ax+by+c=0$
is the general form of the equation of a circle.
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Standard form:
$(x-\displaystyle \frac{1}{2})^{2}+(y-0)^{2}=(\frac{1}{2})^{2}$
$(x-\displaystyle \frac{1}{2})^{2}+y^{2}=\frac{1}{4}$
General form:
$x^{2}-x+\displaystyle \frac{1}{4}+y^{2}=\frac{1}{4}$
$x^{2}+y^{2}-x=0$
