Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

Let the radius of the circle be $r$.

Let each chord have fixed length $x$.

The perpendicular from the centre to a chord bisects the chord.

So, for every chord, half chord = $\frac{x}{2}$

Let $d$ be the distance of the midpoint of the chord from the centre.

Using Baudhāyana - Pythagoras theorem:

$$r^2 = d^2 + \left(\frac{x}{2}\right)^2$$

$$\text{So, } d^2 = r^2 - \left(\frac{x}{2}\right)^2$$

Since $r$ and $x$ are fixed, $d$ is also fixed.

Therefore, every midpoint is at the same distance from the centre.

Hence, the midpoint form a circle with the same centre as the original circle.