Of the two circles, one is fixed, while the radius of the other can vary, only its center is fixed.

How do we choose the radius of the variable circle so that the two circles have a common internal tangent, and the point of tangency on the variable circle is as far as possible from the common axis of symmetry of the two circles?
Let the radius of the fixed circle be the unit length, its center at the origin of the coordinate system, the center of the other circle at point $C(c, 0)$, and its radius $r$. Given the conditions, $c>1$ and $0 \leq r \leq c-1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5dfb95afb451a2721984g-1.jpg?height=1516&width=1870&top_left_y=272&top_left_x=152)

Then, for the angle $\varphi$ of one of the internal common tangents, according to the known construction principle,

$$
\sin \varphi=\frac{1+r}{c}, \quad \cos \varphi=\frac{1}{c} \sqrt{c^{2}-(1+r)^{2}}
$$

From this, the distance of the point of tangency $E$ from $O C$ is

$$
t=r \sin \left(90^{\circ}-\varphi\right)=\frac{r}{c} \sqrt{c^{2}-(1+r)^{2}}
$$

This is non-negative, so it suffices to find the maximum of $t^{2}$:

$$
t^{2}=\frac{c^{2}-1}{c^{2}} r^{2}-\frac{2}{c^{2}} r^{3}-\frac{1}{c^{2}} r^{4}
$$

Let's find where its derivative vanishes. This, after rearrangement, is:

$$
\left(-\frac{2 r}{c^{2}}\right)\left(2 r^{2}+3 r-c^{2}+1\right)=0
$$

The first factor satisfies this at $r=0$, but we are not looking for this, since here $t=0$.

The quadratic polynomial in the large parentheses has two distinct roots: at $r_{1}$ and $r_{2}$, since $r_{1} r_{2}=-\left(c^{2}-1\right) / 20$ (and because the coefficient of the quadratic term is positive).

At $r_{2}$, the polynomial increases, so the derivative decreases. Therefore, at this point, $t^{2}$ has a maximum, and with it, $t$ also has a maximum. This solves the problem.