Consider a family of circle which are passing through the point $(-1, 1)$ and are tangent to $x$-axis. If $(h, k)$ are the coordinates of the centre of the circles, then the set of values of $k$ is given by the interval:

Since the circle is tangent to the $x$-axis and has centre $(h,k)$, its radius equals the distance of the centre from the $x$-axis, so the radius is $|k|$. Also, the circle passes through $(-1,1)$, so the distance from $(h,k)$ to $(-1,1)$ must equal the radius. $$\sqrt{(h+1)^2+(k-1)^2}=|k|$$ Squaring, $$ (h+1)^2+(k-1)^2=k^2 $$ $$ (h+1)^2+k^2-2k+1=k^2 $$ $$ (h+1)^2-2k+1=0 $$ $$ 2k=(h+1)^2+1 $$ $$ k=\frac{(h+1)^2+1}{2} $$ Now $(h+1)^2\ge 0$, so $$ k\ge \frac{1}{2} $$ Every such value is attainable by choosing suitable $h$. Comparing with the options, this matches option $\text{A}$.