A nice observation from Futility Closet:

Eye to eye

Draw two circles of any size and bracket them with tangents, as shown.

The chords in blue will always be equal.

I’m hardly going to let that pass by without a proof, now, am I? Spoilers below the line.

My proof

  • Definitions
    • Let the distance between the centres be $D$.
    • Let the radius of the left-hand circle be $r$ and that of the right-hand circle be $R$. Let their respective centres be $c$ and $C$.
    • Let the length of the left-hand blue line be $2h$ and that of the right-hand blue line be $2H$. (I’ve put the 2 in because I’m going to halve them shortly.)
    • Let the upper points of tangency be $t$ on the left circle and $T$ on the right circle.
    • Let the upper points where the blue chords meet their circles be $x$ on the left and $X$ on the right.

Eye-to-eye diagram

  • Observations
    • The left-hand blue chord crosses $cC$ at right angles; let the point where it does this be $p$.
    • Similarly for the right-hand blue chord; let its crossing-point on $cC$ be $P$.
    • Triangle $cxp$ is right-angled at $p$, and is similar to triangle $cCT$ - so $\frac{h}{r} = \frac{R}{D}$ [1].
    • Similarly, $CXP$ is right-angled at $P$, and is similar to $CcT$ - so $\frac{H}{R} = \frac{r}{D}$ [2].
  • Conclusion
    • From [1], $h = \frac{rR}{D}$
    • From [2], $H = \frac{rR}{D}$
    • So $h=H$ and the two chords are equal.

* Edited 2020-08-17 to improve the diagram and change some colours. Thanks, Barney!