Question

Let the circles $C_1:x^2+y^2=9$ and $C_2:(x-3{)}^2+(y-4{)}^2=16,$ intersect at the points $X$ and $Y.$ Suppose that another circle $C_3:(x-h{)}^2+(y-k{)}^2=r^2$ satisfies the following conditions:

$\left(i\right)$ centre of $C_3$ is collinear with the centres of $C_1$ and $C_2.$

$\left(ii\right)$$C_1$ and $C_2$ both lie inside $C_3$, and

$\left(iii\right)$ $C_3$ touches $C_1$ at $M$ and $C_2$ at $N$

Let the line through $X$ and $Y$ intersect $C_3$ at $Z$ and $W$, and let a common tangent of $C_1$ and $C_3$ be the tangent to the parabola $x^2=8\alpha y.$
There are some expressions given in $List-I$ whose values are given in $List-II$ below:

$\begin{array}{cccc}& \text{List}I& & \text{List}II\\ \left(I\right)& 2h+k& \left(P\right)& 6\\ \left(II\right)& \frac{\text{length of}ZW}{\text{length of}XY}& \left(Q\right)& \sqrt{6}\\ \left(III\right)& \frac{\text{Area of triangle}MZN}{\text{Area of triangle}ZMW}& \left(R\right)& \frac{5}{4}\\ \left(IV\right)& \alpha & \left(S\right)& \frac{21}{5}\\ & & \left(T\right)& 2\sqrt{6}\\ & & \left(U\right)& \frac{10}{3}\end{array}$
Which of the following is the only CORRECT combination?

• A. $\left(I\right),\left(S\right)$
• B. $\left(I\right),\left(U\right)$
• C. $\left(II\right),\left(Q\right)$
• D. $\left(II\right),\left(T\right)$

Answer 1

The correct option is C $\left(II\right),\left(Q\right)$


Centre of $C_1,C_2,C_3$ are collinear and $C_1,C_2$ both lie inside $C_3.$
$MN$ is diameter of $C_3$
$\Rightarrow MN=M{C}_1+C_1{C}_2+C_{2}N$
$\Rightarrow 2r=3+\sqrt{3^2+4^2}+4=12$
$\Rightarrow r=6$
where, $r$ is the radius of $C_3$

Suppose centre of $C_3$ be $(h,k)=(0+r_4\cos \theta ,0+r_4\sin \theta )$$\because r_4=C_1{C}_3=3\text{and}\tan \theta =\frac{4}{3}$
$\therefore C_3=(\frac{9}{5},\frac{12}{5})=(h,k)$
$\therefore 2h+k=6$
Equation of $ZW$ and $XY$ is common chord of circle $C_1=0$ and $C_2=0$


Equation of $ZW:$ $C_1-C_2=0$
$\therefore 3x+4y=9$
Perpendicular distance from centre of $C_3\text{to}ZW=\left|\frac{3\cdot \frac{9}{5}+4\cdot \frac{12}{5}-9}{\sqrt{3^2+4^2}}\right|=\frac{6}{5}$
Length of $ZW$ is,
$ZW=2\sqrt{r^2-(ZW{)}^2}=2\sqrt{6^2-{\left(\frac{6}{5}\right)}^2}=\frac{24}{5}\sqrt{6}$


Perpendicular distance from centre of $C_1\text{to}XY=\left|\frac{3\cdot 0+4\cdot 0-9}{\sqrt{3^2+4^2}}\right|=\frac{9}{5}$
Length of $XY=2\sqrt{3^2-(\frac{9}{5}{)}^2}=\frac{24}{5}$
$\therefore \frac{\text{Length of}ZW}{\text{Length of}XY}=\sqrt{6}$