A circle $\omega$ centered at $K$, $(0.5,0.5)$ is tangent to the $x$- and $y$-axes. Consider the function $$f(x):=\frac{8}{4+x^2} \; \forall x\in\mathbb{R}$$ and a point $P\in f$ such that the distance between $\omega$ and $P$ is minimal.
Determinate the coordinates of $P$.
My attempt so far:
It's trivial that the radius of $\omega$ is $\frac{1}{2}$ which might be proven for instance through reductio ad absurdum and the Pythagorean theorem.
Now, it is a well-known fact, that the Euclidean distance $d$ between $\omega$ and some point $Q(x_q |f(x_q))$ is $$d=\rvert \sqrt{\big(x_q-\frac{1}{2}\bigr)^2+\big(f(x_q)-\frac{1}{2}\bigr)^2}-\frac{1}{2}\lvert$$
This looks something like this
I've tried to simplify the expression and tried to determinate de zeros of the derivative without success since the expression I came up with was to ugly to work with. So, if you've achieved to solve the problem this way 'nicer', I would appreciate if you could show me how.
Anyhow, I was wondering if there is a nicer way to approach this problem, rather than the analytical one, which might be the case since this exercise is though for 15-years-old students...
Thanks in advanced
I feel a bit blocked because I don't know what your students can use... Even though the following ideas could help.
Since the circle and the curve do not intersect:
The distance $d$ is minimal $\Longleftrightarrow d+\frac{1}{2} = \sqrt{\left(x-\frac{1}{2}\right)^2+\left(f(x)-\frac{1}{2}\right)^2}$ is minimal,
and it is $\quad\quad\quad\quad\quad\quad\quad\Longleftrightarrow \left(x-\frac{1}{2}\right)^2+\left(f(x)-\frac{1}{2}\right)^2$ is minimal.
A point $P$ located on the curve $\mathcal{C}_f$ at a minimal distance to $\omega$ lies on a circle $\rho,$ centered at $K$ and tangent to $\mathcal{C}_f$ at $P.$ Thus the line $KP$ is normal to the curve at $P.$
Putting together: $P$ lies at a circle centered at $K,$ which has only one common point with $\mathcal{C}_f.$