Step-by-step Solution:

1. Understand the geometric properties of the circle. The problem states that the circle touches the y-axis at the point (0, 2).

   • For any circle that is tangent to the y-axis at a point (0, k), its center must have the y-coordinate k. So, the center of our circle is of the form C(h, 2).
   • The radius r of the circle is the perpendicular distance from the center (h, 2) to the y-axis (the line x=0). This distance is |h|. So, r = |h|.

2. Formulate the equation of the circle. The standard equation of a circle with center (h, k) and radius r is: $(x - h)^2 + (y - k)^2 = r^2$ Substituting k = 2 and r = |h| (which means $r^2 = h^2$), we get the equation for our specific circle: $(x - h)^2 + (y - 2)^2 = h^2$

3. Use the given point to find the value of h. The circle passes through the point (-1, 0). We can substitute x = -1 and y = 0 into the circle's equation to solve for h. $(-1 - h)^2 + (0 - 2)^2 = h^2$ $(h + 1)^2 + (-2)^2 = h^2$ Expanding the terms: $(h^2 + 2h + 1) + 4 = h^2$ $h^2 + 2h + 5 = h^2$ Subtracting $h^2$ from both sides: $2h + 5 = 0$ $h = -{5 \over 2}$

4. Write the final equation of the circle. Now we have the parameters of the circle:

   • Center: C(h, k) = (-5/2, 2)
   • Radius: r = |h| = |-5/2| = 5/2 The equation of the circle is: $(x - (-5/2))^2 + (y - 2)^2 = (5/2)^2$ $(x + 5/2)^2 + (y - 2)^2 = {25 \over 4}$

5. Check which of the given points lies on the circle. We will substitute the coordinates of each option into the equation of the circle $(x + 5/2)^2 + (y - 2)^2 = 25/4$.

   • A: $\left( { - {3 \over 0},0} \right)$: This option involves division by zero, which is undefined. The point is not valid.

   • B: $\left( { - {5 \over 2},2} \right)$: Let's substitute x = -5/2 and y = 2. $(-5/2 + 5/2)^2 + (2 - 2)^2 = 0^2 + 0^2 = 0$ Since $0 \neq 25/4$, this point is not on the circle. In fact, this is the center of the circle.

   • C: $\left( { - {3 \over 0},\,{5 \over 2}} \right)$: Similar to option A, this option is invalid due to division by zero.

   • D: (- 4, 0): Let's substitute x = -4 and y = 0. $(-4 + 5/2)^2 + (0 - 2)^2$ $= (-8/2 + 5/2)^2 + (-2)^2$ $= (-3/2)^2 + 4$ $= {9 \over 4} + 4$ $= {9 \over 4} + {16 \over 4}$ $= {25 \over 4}$ The result ${25 \over 4}$ matches the right side of the equation ($r^2$). Therefore, the point (-4, 0) lies on the circle.

Conclusion

The circle passes through the point (-4, 0). So, option D is the correct answer.