Question

If I draw two circles of radius 3 cm and 5 cm with a common centre and draw a line AB such that it is a common chord to both circles. CD = 4.46 cm. Find the distance of the chords from the centre and the length AC.

• A. 2, 2.35
• B. 2.35, 2
• C. 3, 2
• D. 2.16, 2.35

Answer 1

The correct option is A

2, 2.35

Given, CD = 4.46 cm

OE is the distance of centre from AB.
So, $OE\perp CD$
$⟹$ CE = DE = 2.23 cm [Perpendicular from the Center to a Chord Bisects the Chord]

Applying Pythagoras Theorem in $\Delta OEC$,
${OE}^2+{CE}^2={OC}^2$
${OE}^2+{2.23}^2=9$
$\therefore {OE}^2=4$
$\Rightarrow OE=2$

In $\Delta OEA$,
${OE}^2+{AE}^2={OA}^2$
$⟹2^2+{AE}^2=5^2$
$⟹$ AE = 4.58 cm

Also, AC = AE - CE
= 4.58 - 2.23
= 2.35 cm