Background
After visiting a plethora of many other related answers on the site, I couldn't answer myself what might be going on over the following scenario, where a relatively simple task is to be done, yet something odd (I believe) is occurring, and so, I'd like to read what others think might be happening here.
Setup
Considering $\left \{A , C \right \}$ as constants, let $f\left(x\right)=A\,x^2+C$ be a certain parabola, and $P=\left\{0 , 0 \right\}$.
Find the shortest distance from $P$ to $f(x)$.
Progress
Considering the above as an optimization problem, I decided to proceed as follows:
$$d^2=\left(y-k\right)^2+\left(x-h\right)^2$$ $$d^2=\left(A\,x^2+C\right)^2+x^2$$ $$d^2=A^2\,x^4+2\,A\,C\,x^2+x^2+C^2$$ $$\frac{\partial{d^2}}{\partial{x}}=4\,A^2\,x^3+4\,A\,C\,x+2\,x = 0$$ $$\left\{x=\pm{{\sqrt{-2\,A\,C-1}}\over{\sqrt{2}\,A}} , x=0 \right\}$$
Even though inappropriate, partial derivative writing was preferred over usual syntax to avoid symbolic confusion; hopefully it's clear why, without degrading the expression's meaning.
Now, for each of the solutions for when derivative is equal to zero:
$$\left\{d^2=\left({{-2\,A\,C-1}\over{2\,A}}+C\right)^2+{{-2\,A\,C-1 }\over{2\,A^2}} , d^2=C^2 \right\} $$ $$\left\{d^2=-{{4\,A\,C+1}\over{4\,A^2}} , d^2=C^2 \right\} $$ $$\left\{\left| d\right| ={{\sqrt{-4\,A\,C-1}}\over{2\,\left| A \right| }} , \left| d\right| =\left| C\right| \right\} $$
And so, I end up with two solutions, respectively $(1)$ and $(2)$.
The Issue
Attempting to use the above formulas succeed in predicting the extremum values for $d$ for many values, like...
But then...
What might be going on? Is there something wrong? Is this expected?
From my point of view, the blue circle should either not exist at all (division by zero), or at least be the same size as the red circle, since blue circle (more precisely, its radius) isn't contributing any useful/true information towards solving the problem stated under these specific values, failing at its basic purpose.
Blue circle obeys $(1)$, while red circle obeys $(2)$.
The first $x$ solution only exist in $\mathbb{R}$ if: $$ 2AC+1<0 $$ $$ AC<-\frac{1}{2} $$
If both solutions exist, you get the minimum of $d^2$ for the first solution, and you can probably prove that for $x=0$ you have a saddle point or local maxima.
If the first one doesn't exist according to the condition, you get a minima at $x=0$.
You always have to find the type of extrema for each value of $x$.