As the circle touches the x-axis at (0,0) then the centre of the circle must be on the y-axis and the y coordinate of the centre must be equal to the radius. So the centre is $(0,r)$. So the equation is:
Then you want to find the point of intersection of the circle and the straight line. Lets rearrange the straight line:
Multiple by $4^2$ to get rid of the fractions:
Collect and sort by powers of $y$:
Now we expect there to be only one point of intersection so this quadratic should have only one solution. This occurs when the discriminant is zero.
Pull out the $16^2$ term and factorise $576$ in preparation to divide out common factor.
Divide by common factor of $256$.
Expand and multiply:
Divide by 4:
So $r=3$ or $r=-12$
Note that $r=-12$ is a valid solution. It means that the circle is below the axis where earlier we assumed it was above the axis (when we set the centre to $(0,r)$).
So the two solutions are: $x^2+(y-3)^2=3^2$ and $x^2+(y+12)^2=12^2$
Or in expanded form: $x^2+y^2-6y=0$ and $x^2+y^2+24y=0$