Equation and Points on Circle Tangent to Axis

To find the equation of the circle, let its center be $(h, k)$ and its radius be $r$.

1. Condition for touching the $y$-axis:
   Since the circle touches the $y$-axis (the line $x = 0$) at the point $(0, 2)$, the $y$-coordinate of the center must be:
   $k = 2$
   The radius of the circle is the perpendicular distance from the center $(h, 2)$ to the $y$-axis:
   $r = |h|$

   Therefore, the equation of the circle can be written as:
   $(x - h)^2 + (y - 2)^2 = h^2$
   Expanding the equation, we get:
   $x^2 - 2hx + (y - 2)^2 = 0$

2. Passing through the point $(-1, 0)$:
   Substituting the coordinates $x = -1$ and $y = 0$ into the equation:
   $(-1)^2 - 2h(-1) + (0 - 2)^2 = 0$
   $1 + 2h + 4 = 0$
   $2h + 5 = 0 \implies h = -\frac{5}{2}$

3. Equation of the circle:
   Substituting $h = -\frac{5}{2}$ back into the circle's equation gives:
   $x^2 + 5x + (y - 2)^2 = 0$
   $x^2 + y^2 + 5x - 4y + 4 = 0$

4. Checking the options:
   Let us test the given points in the circle's equation:

   • For option (D) $(-4, 0)$:
     $(-4)^2 + 5(-4) + (0 - 2)^2 = 16 - 20 + 4 = 0$
     Since the point $(-4, 0)$ satisfies the equation, the circle passes through this point.

Hence, the correct option is (D).