Question

Suppose two chords of a circle are unequal in length. Prove that the chord of larger length is nearer to the centre than the chord of smaller length.

Answer 1

In the figure, $O$ is the center of the circle $AB$ and $CD$ are chords. $AB$ is larger and $CD$ is smaller chord . $OP$ and $OQ$ the distance from the chord $AB$ and $CD$ respectively.

To prove : $OP<OQ$

In triangle $APO$;

$A{O}^2=A{P}^2+P{O}^2\dots \dots \left(1\right)$

In triangle $CQO$;

$O{C}^2=C{Q}^2+O{Q}^2\dots \dots \left(2\right)$

$AO=OC$ (radius of the circle)

From (1) and (2)

$A{P}^2+P{O}^2=C{Q}^2+O{Q}^2$

Let $AP=CQ+x$, therefore,

$(CO+X{)}^2+P{O}^2=C{Q}^2+O{Q}^2$

$(CQ{)}^2+2CQX+P{O}^2=C{Q}^2+O{Q}^2$

$O{Q}^2=O{P}^2+2CQX$

$ORO{Q}^2>O{P}^2$

$OQ>OP$

$OP<OQ$

Hence, the chord of larger length is nearer to the centre than the chord of smaller length.