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 $AB$ to circle $C_1$ touches circles $C_2$ and $C_3$ at $P$ and $Q$ respectively. Line $PQ$ 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 $\to$ Q
• B. III $\to \mathrm{U}$
• C. IV $\to \mathrm{P}$
• D. $\mathrm{IV}⟶\mathrm{T}$

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 $C_1:x^2+y^2=81$