If a be the radius of a circle which touches x-axis at the origin, then its equation is

The equation of the circle with centre at $\left(h,k\right)$ and radius equal to $a$ is ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$

When the circle passes through the origin and centre lies on $x-$ axis

$\Rightarrow k=a$ and $h=0$

Then the equation ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$ becomes ${{x}^{2}+\left(y-a\right)}^{2}={a}^{2}$

If a circle passes through the origin and centre lies on $x-$axis then the abscissa will be equal to the radius of the circle and the $y-$co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form

${\left(y\pm a\right)}^{2}+{x}^{2}={a}^{2}$

$\Rightarrow {x}^{2}+{a}^{2}\pm 2ay+{y}^{2}={a}^{2}={x}^{2}+{y}^{2}\pm 2ay=0$ is the required equation of the circle.