Question

3 circles $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ with centres $\mathrm{O}_1, \mathrm{O}_2$ and $\mathrm{O}_3$ of radius 9,6 and 3 respectively touches each other as shown in figure. $A$ chord $A B$ to circle $C_1$ touches circles $C_2$ and $C_3$ at $P$ and $Q$ respectively. Line $P Q$ meets line joining centres of $\mathrm{C}_2$ and $\mathrm{C}_3$ at $\mathrm{T}, \mathrm{D}$ is mid-point of $\mathrm{Q}$ and $\mathrm{O}_2$.

List I		List II
I	Length of chord PQ is	P	$3 \sqrt{2}$
II	Length of chord AB is	Q	$4 \sqrt{14}$
III	Length of TP is	R	$6 \sqrt{2}$
IV	Length of $O_3 D$ is	S	$4 \sqrt{7}$
		T	$2 \sqrt{14}$
		U	$12 \sqrt{2}$

Which of the following is the only incorrect combination?

• A. II $\rightarrow$ Q
• B. III $\rightarrow \mathrm{U}$
• C. IV $\rightarrow \mathrm{P}$
• D. $\mathrm{IV} \longrightarrow \mathrm{T}$

Answer 1

Answer: (C)

Length of $\mathrm{PQ}=\mathrm{TCT}=\sqrt{9^2-(6-3)^2}=6 \sqrt{2}$
let $C_1: x^2+y^2=9^2$
let $\angle \mathrm{QTN}=\theta$
$\therefore \frac{\mathrm{NT}}{\mathrm{MT}}=\frac{\mathrm{r}_3}{\mathrm{r}_2}=\frac{3}{6}=\frac{1}{2}=\frac{\mathrm{TQ}}{\mathrm{TP}} \Rightarrow \mathrm{N}$ is midpoint of TM
$ \Rightarrow \mathrm{TP}=2 \mathrm{TQ} \Rightarrow \mathrm{TQ}=\mathrm{PQ} $
$\Rightarrow \mathrm{TQ}=6 \sqrt{2} $
$ \therefore \mathrm{NT}=\sqrt{72+9}=9 $
$ \Rightarrow \mathrm{LT}=9+(9-3)=15 $
$ \because \mathrm{L}(0,0) \Rightarrow \mathrm{T}(-15,0) $
$ \sin \theta=\frac{\mathrm{NQ}}{\mathrm{NT}}=\frac{3}{9}=\frac{1}{3} $
$ \Rightarrow \cos \theta=\frac{6 \sqrt{2}}{9}=\frac{2 \sqrt{2}}{3}$
Any point on chord AB be $\left(-15+\frac{2 \sqrt{2} \mathrm{r}}{3}, \frac{\mathrm{r}}{3}\right)$ lie on circle $C_1: x^2+y^2=81$