The circle passing through the points $(-1,0)$ and touching the y-axis at $(0,2)$ also passes through the point:

Let $\left( h, k \right)$ be center of circle.

Circle touches the $y$-axis.

$\therefore$ Radius of circle $= h$

Equation of circle-

${\left( x - h \right)}^{2} + {\left( y - k \right)}^{2} = {h}^{2} ..... \left( 1 \right)$

Since the circle passes through $\left( 0, 2 \right)$

Therefore,

${h}^{2} + {\left( 2 - k \right)}^{2} = {h}^{2}$

$\Rightarrow {\left( k - 2 \right)}^{2} = 0$

$\Rightarrow k = 2$

Given that the circle also passes through $\left( -1, 0 \right)$,

Therefore,

${\left( -1 - h \right)}^{2} + {2}^{2} = {h}^{2}$

${h}^{2} + 1 + 2h + 4 = {h}^{2}$

$h = - \cfrac{5}{2}$

Substituting the value of $h$ and $k$ in ${eq}^{n} \left( 1 \right)$, we get

${\left( x + \cfrac{5}{2} \right)}^{2} + {\left( y - 2 \right)}^{2} = {\left( \cfrac{5}{2} \right)}^{2}$

Now, we can find the point from which the circle passes.

As the point $\left( 4, 0 \right)$ satisfy the equation of circle.

Thus, the circle will also pass through the point $\left( -4, 0 \right)$.

Hence the correct answer is $\left( D \right) \left( -4, 0 \right)$.