Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.

Solution:

Given that,

Radius of the circle (OA) = 6 cm

Distance (OC) = 4 cm

In ΔOCA, by Pythagoras theorem

$\mathrm{AC}^{2}+\mathrm{OC}^{2}=\mathrm{OA}^{2}$

$\Rightarrow \mathrm{AC}^{2}+4^{2}=6^{2}$

$\Rightarrow \mathrm{AC}^{2}=36-16$

$\Rightarrow \mathrm{AC}^{2}=20$

$\Rightarrow \mathrm{AC}=\sqrt{20}$

⟹ AC = 4.47 cm

We know that the perpendicular distance from centre to chord bisects the chord.

AC = BC = 4.47 cm

Then AB = 4.47 + 4.47

= 8.94 cm