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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!

Sketch

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  • $\begingroup$ $d=R-\arcsin(\sqrt{(1-R)^2-\cos^2 \theta})$ $\endgroup$
    – N74
    Jun 21, 2016 at 18:58
  • $\begingroup$ I'm not sure if it is $\sin^2 \theta$ instead of $\cos^2 \theta$, but you can verify. If tomorrow nobody answered, I'll give you the demonstration. $\endgroup$
    – N74
    Jun 21, 2016 at 19:00

2 Answers 2

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

enter image description here

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.

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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 $\ell/2 - x$.
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  • $\begingroup$ Thanks, you wish to say that measure of angle $ABO$ is $\frac{\pi}{2} - \theta$, not $2\pi - \theta$, right? $\endgroup$
    – Alex
    Jun 21, 2016 at 19:22
  • $\begingroup$ Oops, yes. Fixed! $\endgroup$
    – user326210
    Jun 21, 2016 at 19:32
  • $\begingroup$ It looks like there's a flaw in step 3. The points $O$, $P$ and $A'$ are not necessarily collinear, so you cannot say that length of $OA'$ is $1-R$, surely? $\endgroup$
    – Alex
    Jun 21, 2016 at 20:02
  • $\begingroup$ @Alex I might have been mistaken, but I thought that they must be collinear. It seemed to me that since both circles share a tangent (osculate) at point $P$, then a line drawn from $O$ to $P$ meeting the shared tangent perpendicularly will also pass through the center of the relocated small circle, $A^\prime$. $\endgroup$
    – user326210
    Jun 21, 2016 at 20:57

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