How to find the tangency condition for this circle geometry problem?

Question

Suppose I have a circle $C$ of radius $1$, and I have a chord of this circle, of given length $l$. The chord makes a known angle $\theta$ with the tangent to the circle. I position a smaller circle $C_{R}$ of radius $R<1$ on the $\textit{midpoint}$ of the chord. Then I move the small circle by some distance $d$ $\textit{along}$ (parallel to) the chord.

For a given $R$, $l$ and $\theta$, I would like to find the maximum distance $d$ I can move $C_{R}$ by such that $C_{R}$ still stays inside the larger circle $C$: this happens when $C_{R}$ is just $\textit{tangent}$ to $C$.

How would I find this $d$ in terms of the given quantities? Any help would be appreciated!

Answer 1

As mentioned, the limiting case is when the small circle touches the larger circle internally. Then, we have the figure below:-

Applying Pythagoras theorem to the red triangle, we have $OM^2 = 1 – (o.5L)^2$.

Applying Pythagoras theorem to the green triangle, we have $(R + d)^2 = (1 – R)^2 – OM^2$

Eliminating $OM^2$ from the two equations, we get $d$ in terms of $R$ and $L$ only. $\theta$ is not included because $L$ and $\theta$ are inter-related.

Answer 2

Here's how I think you solve the problem. Label the center of the unit circle $O$, the midpoint $A$ of the chord, a point $B$ where the chord meets the unit circle, and pick a point $T$ on the line tangent to the unit circle at point $B$. After you've moved the tiny circle, call its new center $A^\prime$.

You want to find the length $\overline{AA^\prime}$.

1. You know that $\overline{OB}$ is a radius of the unit circle, with length 1.
2. You know that $\angle OBT$ is a right angle, since $\overline{BT}$ is tangent to the circle. You also know that the measure of $\angle ABT$ is $\theta$. Angle $\angle ABO$ is complementary and therefore has measure $(\pi/2)-\theta$.
3. After you move the tiny circle, it touches the bigger circle at some point $P$, so $\overline{OP}$ is a radius of the unit circle, with length 1; $\overline{A^\prime P}$ is a radius of the smaller circle, with length $R$. Hence the length of $\overline{OA^\prime}$ is $1-R$.
4. Now, consider triangle $\Delta OA^\prime B$. You know the length of two sides and an angle not included in it. So you can solve for the length of $\overline{A^\prime B}$— call it $x$.
5. Since $\overline{AB}$ is half the chord, it has length $\ell/2$. Hence the overall distance between $A$ and $A^\prime$ is just $\ell/2 - x$.