Not sure why this value arises when I am trying to find the tangential intersection? if we consider the circle, whose equation is given by $$x^2+(y-2)^2=1$$
and the parabola $$y=kx^2$$ 
We wish to find the values of $k$ for which the parabola will touch the circle (not intersect but touch). 
Current Solution
Let us simply substitute for $x^2=\frac{y}{k}$ into the circle, this gives: 
$$\frac{y}{k}+y^2-4y+4=1 \Rightarrow ky^2+(1-4k)y+3k=0 $$
Now evaluating $$\Delta=0$$
we are solving for the values of $k$ for which the parabola only intersects the circle once. 
And so we arrive at $$\Delta=(1-4k)^2-12k=1-8k+16k^2-12k^2=4(k^2-2k)+1$$
$$\Delta=4(k-1)^2-3=0\Rightarrow k=\frac{2\pm\sqrt{3}}{2}$$
However, upon further inspection, the specific value $$k=\frac{2-\sqrt{3}}{2}$$
causes the parabola to never intersect the circle, and thus we only take the other value that we found.
My Problem 
While this question is "solved", I do not know why such a value of $k$ rose in the first place? Am I interpreting the discriminant incorrectly? 
Because such a value for $k$ will make the parabola touch the circle twice due to of course, the symmetry of this particular example. 
 A: The step which conceals the introduction of additional, ineligible roots is this one:

Now evaluating $$\Delta=0$$
  we are solving for the values of $k$ for which the parabola only intersects the circle once. 

The problem here is that $\Delta = 0$ guarantees a root of multiplicity (at least) $2$ to the system of equations, but it does not guarantee that the double root is real. Indeed, the second value of $k$ does in fact correspond to a double root, only that the respective root is complex, outside the real plane, so it doesn't count as an intersection of the real conics.
To follow what happens step by step, take the value $k=\frac{2-\sqrt{3}}{2}$ and replace it back into the equation $ky^2+(1-4k)y+3k=0$. Since it's already established that $\Delta=0$ the double root $y$ is given by:
$$
y = \frac{1-4k}{2k}=\frac{3 - 2 \sqrt{3}}{2 - \sqrt{3}}=(3 - 2 \sqrt{3})(2 + \sqrt{3}) = -\sqrt{3}
$$
Then $y= k x^2$ reduces to $x^2 = \frac{y}{k} = \frac{-2\sqrt{3}}{2-\sqrt{3}} = -2\sqrt{3}(2+\sqrt{3}) = -2(3 + 2 \sqrt{3}) \lt 0$ which has no real roots. The complex roots $x=\pm i\,\sqrt{2(3 + 2 \sqrt{3})}$ do in fact correspond to the double roots of the two equations, but they are not "intersections" in the real plane.
A: The problem is that in your substitution, we must require $x^2=\frac{y}{k}\ge0$ and thus:
$${y\over k}=\frac{4k-1\pm\sqrt{4(k^2-2k)+1}}{2k^2}\ge0$$
When $k=\frac{2-\sqrt3}{2}$, $\Delta=0$ so we have one solution but: $$x^2={y\over k}=\frac{4-2\sqrt3-1}{2({2-\sqrt3\over2})^2}<0$$
it is negative hence this solution must be rejected. $\Delta=0$ doesn't ensure $y\ge0$.
A: To find the "lost" solution for $k=(2-\sqrt{3})/2$, plot your curves on a coordinate plane where $x$ is pure imaginary instead of real, while $y$ is still real.  What appears as a circle in the all-real plane is now a hyperbola, and the parabola bends down rather than up to become tangent to this hyperbola.
So, the "lost" solution involves imaginary coordinates.
