Question

If I draw two circles of radius 3 cm and 5cm with common centre and draw a line AB such that it is a chord to both the circles and length of CD is $2\sqrt{5}$ cm, then find the distance of the chords from the centre and the length AC.

• A. 2 cm, 2.346 cm
• B. 2.35 cm, 2 cm
• C. 3 cm, 2 cm
• D. 2.16 cm, 2.346 cm

Answer 1

The correct option is A

2 cm, 2.346 cm

Given, CD = $2\sqrt{5}$ cm
Since $OE\perp CD,CE=DE=\sqrt{5}\text{cm}$.
(Perpendicular from the centre to a chord bisects the chord.)
Apply Pythagoras' theorem in $\Delta OEC$. We get
${OE}^2+{CE}^2={OC}^2$
${OE}^2+{\left(\sqrt{5}\right)}^2=9$ [OC = Radius of smaller circle = 3 cm]
$⟹{OE}^2=9-5$
$\therefore {OE}^2=4$
$\Rightarrow OE=2$ cm
In $\Delta OEA$,
${OE}^2+{AE}^2={OA}^2$
$⟹2^2+{AE}^2=5^2$ [OA = Radius of bigger circle = 5 cm]
$⟹AE=\sqrt{21}\text{cm}$
Now,
$AC=AE-CE=\sqrt{21}-\sqrt{5}$
$=4.582-2.236$
$=2.346\text{cm}$