Question

In the given figure points P and Q are the centres of the circles, Radius QN = 3, PQ = 9. M is the point of contact of the circles. Line ND is tangent to the larger circle. Point C lies on the smaller circle .Detremine NC, ND and CD.

Answer 1

Construction: Join PD and MC.


QM = NQ = 3 units (Radius of the smaller circle)
PM = PQ $-$ QM = 9 $-$ 3 = 6 units
PD = 6 units (PM=PD)
PN = NQ + PQ = 9 + 3 = 12 units
$$\begin{gathered} \angle \mathrm{NDP}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \mathrm{So},\mathrm{in}\mathrm{right}∆\mathrm{NDP},\mathrm{we}\mathrm{have}: \\ \mathrm{ND}=\sqrt{{\mathrm{PN}}^2-{\mathrm{PD}}^2} \\ =\sqrt{{12}^2-6^2} \\ =\sqrt{144-36} \\ =\sqrt{108} \\ =6\sqrt{3}\mathrm{units} \end{gathered}$$

$\angle$MCN = 90$°$ (Angle in a semicircle is a right angle)
$\angle$MCN = $\angle$PDN = 90$°$
So, the corresponding angles are equal.
Hence, $\mathrm{CM}\parallel \mathrm{PD}$
∴ $\angle$NMC = $\angle$NPD
$$\begin{gathered} \mathrm{In}△\mathrm{NCM}\mathrm{and}△\mathrm{NPD},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{MCN}=\angle \mathrm{PDN}=90°(\mathrm{Corresponding}\mathrm{angle}) \\ \angle \mathrm{NMC}=\angle \mathrm{NPD}(\mathrm{Corresponding}\mathrm{angle}) \\ △\mathrm{NCM}~△\mathrm{NDP}(\mathrm{By}\mathrm{AA}\mathrm{similarity}) \\ \mathrm{Now},\frac{\mathrm{NC}}{\mathrm{ND}}=\frac{\mathrm{NM}}{\mathrm{NP}} \\ \Rightarrow \frac{\mathrm{NC}}{6\sqrt{3}}=\frac{6}{12} \\ \Rightarrow \mathrm{NC}=3\sqrt{3}\mathrm{units} \end{gathered}$$
$\therefore \mathrm{CD}=\mathrm{ND}-\mathrm{NC}=6\sqrt{3}-3\sqrt{3}=3\sqrt{3}\mathrm{units}$