Question

Two circles of radius 3 cm and 5 cm have a common centre, O. AB is a chord to both the circles and length of CD is $2\sqrt{5}$ cm. Find the distance of the chord from the centre and the length AC.

[3 Marks]

Answer 1

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.) [1 Mark]

Apply Pythagoras' theorem in $\Delta OEC$. We get
${OE}^2+{CE}^2={OC}^2$
${OE}^2+{\left(\sqrt{5}\right)}^2=9$
$⟹{OE}^2=9-5$
$\therefore {OE}^2=4$
$\Rightarrow OE=2cm$ [1 Mark]
In $\Delta OEA$,
${OE}^2+{AE}^2={OA}^2$
$⟹2^2+{AE}^2=5^2$
$⟹AE=\sqrt{21}\text{cm}$
Now,
$AC=AE-CE=\sqrt{21}-\sqrt{5}$
$=4.582\text{cm}-2.236\text{cm}$
$=2.346\text{cm}$ [1 Mark]