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

Let the centre of a required circle be $$(h,k)$$ and its radius be $$r$$.

Because the circle is tangent to the $$x$$-axis, the perpendicular distance from the centre to the $$x$$-axis equals the radius: $$r = |k|$$.

The circle also passes through the point $$(-1,1)$$, so the distance from this point to the centre equals the radius:
$$\sqrt{(h+1)^2 + (k-1)^2} = |k| \; -(1)$$

Square both sides of $$(1)$$:
$$(h+1)^2 + (k-1)^2 = k^2$$

Expand and simplify:
$$(h+1)^2 + k^2 - 2k + 1 = k^2 \quad\Longrightarrow\quad (h+1)^2 = 2k - 1 \; -(2)$$

The left-hand side of $$(2)$$, being a square, is always $$\ge 0$$. Hence, $$2k - 1 \ge 0 \;\Longrightarrow\; k \ge \tfrac{1}{2}$$.

Conversely, for every $$k \ge \tfrac{1}{2}$$ we can choose $$h$$ to satisfy $$(h+1)^2 = 2k-1$$, so a circle with those parameters does exist.
(For $$k = \tfrac{1}{2}$$ we get $$h = -1$$, giving the circle with centre $$(-1,\tfrac12)$$ and radius $$\tfrac12$$.)

Therefore the possible values of $$k$$ form the interval $$k \ge \tfrac12$$.

Option B which is: $$k \ge 1/2$$