The locus of the centres of the circles, which touch the circle, $$x^2 + y^2 = 1$$ externally, also touch the y-axis and lie in the first quadrant, is:

Let us denote the required variable circle by $$S$$.

We write its centre as $$C(h,k)$$ and its radius as $$r$$. Because the locus is required only in the first quadrant, we shall always have $$h \gt 0$$ and $$k \gt 0$$.

First condition: “the circle touches the $$y$$-axis.” The $$y$$-axis is the line $$x = 0$$, and the (perpendicular) distance of the centre $$C(h,k)$$ from this line is simply $$|h|$$. Touching the $$y$$-axis means that this distance equals the radius. Hence

$$r = h \quad\text{(because }h\gt 0\text{)}.$$

Second condition: “the circle touches the circle $$x^{2}+y^{2}=1$$ externally.” The given circle has centre $$O(0,0)$$ and radius $$1$$. For two circles to touch externally, the distance between their centres equals the sum of their radii. So, by the distance formula, we must have

$$\sqrt{(h-0)^{2}+(k-0)^{2}} \;=\; 1 + r.$$

Simplifying the left‐hand side and now substituting the value $$r = h$$ from the first condition, we get

$$\sqrt{h^{2}+k^{2}} \;=\; 1 + h.$$

To remove the square root we square both sides:

$$h^{2}+k^{2} \;=\; (1+h)^{2}.$$

Expanding the right‐hand side gives

$$h^{2}+k^{2} \;=\; 1 + 2h + h^{2}.$$

We observe that the term $$h^{2}$$ occurs on both sides; subtracting $$h^{2}$$ from each side leaves

$$k^{2} \;=\; 1 + 2h.$$

Replacing $$h$$ by $$x$$ and $$k$$ by $$y$$—because every point $$C(h,k)$$ on the locus is an ordinary point $$(x,y)$$ in the coordinate plane—we obtain the Cartesian equation of the locus:

$$y^{2} \;=\; 1 + 2x.$$

Finally, since the centres lie in the first quadrant, we keep the positive square root of $$y^{2}$$ and also insist on $$x \ge 0$$. Therefore

$$y \;=\; \sqrt{\,1 + 2x\,}, \quad x \ge 0.$$

Hence, the correct answer is Option A.