Question

Answer the following by appropriately matching the lists based on the information given in the paragraph.

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:

(i) centre of $C_3$ is collinear with the centres of $C_1$ and $C_2$

(ii) $C_1$ and $C_2$ both lie inside $C_2$, and

(iii) $C_3$ 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 a tangent to the parabola $x^2=8\alpha y$.

There are some expressions given in the $List–I$ whose values are given in $List–II$ below:

$List-I$	$List-II$
$\left(I\right)2h+k$	$\left(P\right)6$
$\left(II\right)\frac{LengthofZW}{LengthofXY}$	$\left(Q\right)\sqrt{6}$
$\left(III\right)\frac{AreaoftriangleMZN}{AreaoftriangleZMW}$	$\left(R\right)\frac{5}{4}$
$\left(IV\right)\alpha$	$\left(S\right)\frac{21}{5}$
	$\left(T\right)2\sqrt{6}$
	$\left(U\right)\frac{10}{3}$

Which of the following is the only INCORRECT combination?

• A. $\left(IV\right),\left(S\right)$
• B. $\left(IV\right),\left(U\right)$
• C. $\left(III\right),\left(R\right)$
• D. $\left(I\right),\left(P\right)$

Answer 1

The correct option is A

$\left(IV\right),\left(S\right)$

Step 1. Draw the diagram according to the question:

Given Centers are $C_1,C_2,C_3$ are collinear

$\Rightarrow$$\left|\begin{array}{ccc}0& 0& 1\\ 3& 4& 1\\ h& k& 1\end{array}\right|=0$

$\Rightarrow$ $3k=4h$ ….(1)

$MN$ is Diameter of $C_3$

$\Rightarrow$$\begin{array}{rcl}MN& =& 3+\sqrt{\left(3-0\right)+{\left(4-0\right)}^2}+4\\ & =& 12\end{array}$

$\Rightarrow$ $r=6$ ….(2)

Also, Given $C_3$ touches $C_1$ at $M$

Where, $C_1\left(0,0\right)$ and $C_3=\left(h,k\right)$

$\Rightarrow$$\left|C_1{C}_3\right|=\left|r-3\right|$

$\Rightarrow$$h^2+k^2=9$ ….(3)

Step 2. From equation (1) and (3), we get

$h=\pm \frac{9}{5}$ and $k=\pm \frac{12}{5}$

So, Centre of $C_3$ is $\left(\frac{9}{5},\frac{12}{5}\right)$

Now, equation of $XY$ is

$C_1-C_2=0$

$\Rightarrow$$6x+8y=18$

$\Rightarrow$$3x+4y=9$ ….(4)

Now, $C_{1}P=\frac{9}{5}($distance from origin to equation of $XY)$

$\begin{array}{rcl}P{Y}^2& =& C_1{Y}^2-C_1{P}^2\\ & =& 9-\frac{81}{25}\\ & =& \frac{144}{25}\end{array}$

$\begin{array}{rcl}XY& =& 2PY\\ & =& 2\times \frac{12}{5}\\ & =& \frac{24}{5}\end{array}$

Similarly, equation of $ZW$ is $3x+4y=9$

$\therefore$Length of perpendicular from $C_3$ to $ZW$ $=\left|\frac{3\left({\displaystyle \frac{9}{5}}\right)+4\left({\displaystyle \frac{12}{5}}\right)-9}{5}\right|$

$=\frac{6}{5}$

Now,

$\begin{array}{rcl}ZW& =& 2\sqrt{{\left(6\right)}^2-{\left(\frac{6}{5}\right)}^2}\\ & =& \frac{24\sqrt{6}}{5}\end{array}$

Step 3. Solve Every option one by one, we get

(A)

$\begin{array}{rcl}2h+k& =& 2\times \frac{9}{5}+\frac{12}{5}\\ & =& \frac{30}{5}\\ & =& 6\end{array}$

$\mathbf{\therefore }\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }\mathbf{\left(}\mathit{P}\mathbf{\right)}$

(B)

$\frac{LengthofZW}{LengthofXY}=\sqrt{6}$

$\mathbf{\therefore }\mathbf{\left(}\mathit{I}\mathit{I}\mathbf{\right)}\mathbf{\to }\mathbf{\left(}\mathit{Q}\mathbf{\right)}$

(C)

$\begin{array}{rcl}\frac{\mathbf{A}\mathbf{r}\mathbf{e}\mathbf{a}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{t}\mathbf{r}\mathbf{i}\mathbf{a}\mathbf{n}\mathbf{g}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{M}\mathbf{Z}\mathbf{N}}{\mathbf{A}\mathbf{r}\mathbf{e}\mathbf{a}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{t}\mathbf{r}\mathbf{i}\mathbf{a}\mathbf{n}\mathbf{g}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{Z}\mathbf{M}\mathbf{W}}\mathbf{}\mathbf{}& \mathbf{=}& \mathbf{}\frac{{\displaystyle \frac{1}{2}\times MN\times PN}}{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathbf{Z}\mathbf{W}\mathbf{\times }\mathbf{M}\mathbf{P}}\\ & =& \frac{{\displaystyle \frac{1}{2}}\times 2\times 6\times {\displaystyle \frac{1}{2}}\left(ZW\right)}{{\displaystyle \frac{1}{2}}\left(ZW\right)\times \left(M{C}_1+C_{1}P\right)}\\ & =& \frac{6}{\left(3+{\displaystyle \frac{9}{5}}\right)}\\ & =& \frac{6\times 5}{24}\\ & =& \frac{5}{4}\end{array}$

$\mathbf{\therefore }\mathbf{\left(}\mathit{I}\mathit{I}\mathit{I}\mathbf{\right)}\mathbf{\to }\mathbf{\left(}\mathit{R}\mathbf{\right)}$

(D)

Tangent at $M$, is also tangent to parabola $x^2=8\alpha y$

So, slope of tangent at $M=-\left(\frac{1}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)$

$=-\frac{3}{4}$

So equation of tangent at $M$ to $C_1$ is

$y=mx\pm a\sqrt{1+m^2}$ where $a=3,m=-\frac{3}{4}$

$\Rightarrow$ $y=-\frac{3}{4}x-3\sqrt{1+\frac{9}{16}}$

$\Rightarrow$$4y+3x+15=0$

which is tangent to $x^2=8\alpha y$

So equation of tangent to $x^2=8\alpha y$ is $y=mx-2\alpha m^2$

Step 4. By comparing both equation, we get

$\alpha =\frac{10}{3}$

$\mathbf{\therefore }\mathbf{\left(}\mathit{I}\mathit{V}\mathbf{\right)}\mathbf{\to }\mathbf{\left(}\mathit{U}\mathbf{\right)}$

Hence, Option ‘A’ is Incorrect.