Shortest chord through a fixed interior point: Of all chords of a circle that pass through a given interior point, which chord is the shortest?

Introduction / Context:Fix a point P inside a circle. Among all chords through P, their lengths vary with orientation. We want the orientation that minimizes the chord length.

Given Data / Assumptions:

• Circle with radius R and center O.
• P is a fixed interior point (OP < R).
• We consider all possible chords through P.

Concept / Approach:For any chord, its length is 2√(R^2 − d^2), where d is the perpendicular distance from O to the chord. For chords through P, d varies with orientation. The chord is shortest when d is maximized, i.e., when the chord is as far from the center as possible. That occurs when P is the midpoint of the chord (the chord is “bisected at P”), placing the chord at distance OP from O.

Step-by-Step Solution:If P is the chord’s midpoint, d = OP, giving minimal length L_min = 2√(R^2 − OP^2).Any tilt away from this midpoint position reduces d below OP, increasing chord length.

Verification / Alternative check:Compare with the diameter through P (passing through O): that chord has d = 0, which yields the maximum length, not the minimum.

Why Other Options Are Wrong:“Passes through the centre” gives the longest chord (a diameter). Trisecting at P or vague orientations do not maximize d.

Common Pitfalls:Assuming the diameter through P is always the extremum (it is, but for maximum length).

Final Answer:Is bisected at the point