Question

In the given figure, chord AB is larger than chord CD. The relation between OM and ON is:

Options

• OM = ON

• OM < ON

• OM > ON

• OM + ON = AB

Answer 1

OM < ON

Explanation:

• In a circle of radius r, the perpendicular distance d from the centre O to a chord of length L is given by the Pythagorean relation:
  `d^2 + \left(\frac{L}{2}\right)^2 = r^2 \implies d = \sqrt{r^2 - \left(\frac{L}{2}\right)^2}`
• As the length of a chord L increases, the term subtracted from r² becomes larger, making the perpendicular distance d from the centre smaller (longer chords are closer to the centre).
• Since chord AB > chord CD, the distance OM of chord AB from the centre must be strictly less than the distance ON of chord CD from the centre:
  OM < ON