Question

O is the centre of the given circle, AB and CD are chords such that ON < OM. Which of the following is not correct?

• A.
  $AN>CM$
• B.
  $O{A}^2=O{N}^2+A{N}^2$
• C.
  $C{M}^2=O{C}^2+O{M}^2$
• D.
  $AB>CD$

Answer 1

The correct option is C

$C{M}^2=O{C}^2+O{M}^2$
ON < OM (Given)

$\Rightarrow (ON{)}^2<(OM{)}^2\Rightarrow –(ON{)}^2>–(OM{)}^2\Rightarrow (OA{)}^2–(ON{)}^2>(OC{)}^2–(OM{)}^2(\because OA=OC)\Rightarrow (AN{)}^2>(CM{)}^2\text{(By Pythagoras theorem)}\Rightarrow AN>CM\dots ..\left(i\right)\text{Multiplying (i) by 2, we get}2AN>2CM$

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

$\therefore 2AN=ABand2CM=CD\Rightarrow AB>CD\dots ..\left(ii\right)$

Now, in $\Delta OCM,$ by Pythagoras theorem,

$(OC{)}^2=(OM{)}^2+(CM{)}^2\Rightarrow (OC{)}^2+(OM{)}^2=2(OM{)}^2+(CM{)}^2\ne (CM{)}^2\dots ..\left(iii\right)Thus,C{M}^2\ne O{C}^2+O{M}^2$

Hence, the correct answer is option (c).