Let the circles $C_1:x^2+y^2=9$ and $C_2:{\left(x-3\right)}^2+{\left(y-4\right)}^2=16,$ intersect at the points $X$ and $Y.$ Suppose that another circle $C_3:{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ 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 a tangent to the parabola $x^2=8\alpha y.$
There are some expression given in the List- $\mathrm{I}$ whose values are given in List- $\mathrm{I}\mathrm{I}$ below:
 

	List-  I		List-  II
$\left(I\right)$	$2h+k$	$\left(P\right)$	$6$
$\left(II\right)$	$\frac{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }ZW}{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }XY}$	$\left(Q\right)$	$\sqrt{6}$
$\left(III\right)$	$\frac{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{ }MZN}{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{ }ZMW}$	$\left(R\right)$	$\frac{5}{4}$
$\left(IV\right)$	$\alpha$	$\left(S\right)$	$\frac{21}{5}$
		$\left(T\right)$	$2\sqrt{6}$
		$\left(U\right)$	$\frac{10}{3}$

Which of the following is the only INCORRECT combination?

• Correct answer
  A

  $\left(IV\right)-\left(S\right)$
• B

  $\left(IV\right)-\left(U\right)$
• C

  $\left(III\right)-\left(R\right)$
• D

  $\left(I\right)-\left(P\right)$

Solution

Given centre of $C_1, C_2$ and $C_3$ are collinear hence
$\left|\begin{array}{ccc}0& 0& 1\\ 3& 4& 1\\ h& k& 1\end{array}\right|=0$
$3k=4h$ …(i)
$\Rightarrow MN$ is diameter of $C_3$
$MN=M{C}_1+C_1{C}_2+C_{2}N$
$\begin{array}{cc}2r=r_1+C_1{C}_2+r_2& \left(\because MN=2r\right)\end{array}$
$2r=3+\sqrt{{\left(3-0\right)}^2+{\left(4-0\right)}^2}+4$
$\boxed{r=6}$ …(ii)
$\Rightarrow$ Given $C_3$ touches $C_1$ at $\begin{array}{cc}M& \end{array}$
So $\begin{array}{cc}\left|C_1{C}_3\right|=\left|r-3\right|& \end{array}$
$h^2+k^2=9$ …(iii)
From (i) and (iii)
$h=\pm \frac{9}{5}$ and $k=\pm \frac{12}{5}$
So centre of $C_3$ will be $\left(\frac{9}{5},\frac{12}{5}\right)$
$\Rightarrow$ Now equation of common chord of $C_1$ and $C_2$ will be
$C_1-C_2=0$
$6x+8y=18$
Equation of line $XY$ is $3x+4y=9$ …(iv)
Distance of line $XY$ from origin
$C_{1}P=\frac{9}{5}$
$\Rightarrow$ now in $∆C_{1}Py$
${\left(C_{1}P\right)}^2+{\left(PY\right)}^2={\left(C_{1}Y\right)}^2$
$\begin{array}{cc}\frac{81}{25}+{\left(PY\right)}^2=\left(9\right)& \left(\because C_{1}Y=r_1=3\right)\end{array}$
${\left(PY\right)}^2=\frac{144}{25}\Rightarrow PY=\frac{12}{5}$
Length of $XY=2PY=\frac{24}{5}$
$\Rightarrow$ Line $ZW$ is line $XY$
Equation of $ZW=3x+4y=9$
Distance of $C_3$ from $ZW=\left|\frac{3\left(\frac{9}{5}\right)+4\left(\frac{12}{5}\right)-9}{5}\right|$
$ZW=\frac{6}{5}$
now $ZW=2\sqrt{{\left(6\right)}^2-{\left(\frac{6}{5}\right)}^2}$
$ZW=\frac{24\sqrt{6}}{5}$

$\left(\mathrm{I}\right) 2h+k=2\times \frac{9}{5}+\frac{12}{5}=\frac{30}{5}=6$
$\left(\mathrm{I}\mathrm{I}\right)\frac{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }ZW}{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }XY}=\sqrt{6}$
$\left(\mathrm{I}\mathrm{I}\mathrm{I}\right)\frac{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }∆MZN}{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }∆ZMW}=\frac{\frac{1}{2}\times MN\times PZ}{\frac{1}{2}\times ZW\times MP}$
$=\begin{array}{cc}\frac{\frac{1}{2}MN\left(\frac{1}{2}ZW\right)}{\frac{1}{2}\times ZW\left(MG+GP\right)}& \left(\therefore PZ=\frac{1}{2}ZW\right)\end{array}$
$=\begin{array}{cc}\frac{\frac{1}{2}\times 12\times \frac{12\sqrt{6}}{5}}{\frac{1}{2}\times \frac{24\sqrt{8}}{5}\left(\frac{24}{5}\right)}& \left(\begin{array}{c}\therefore M{C}_1=3\\ \therefore C_{1}A=\frac{9}{5}\end{array}\right)\end{array}$
$=\frac{5}{4}$
$\left(\mathrm{I}\mathrm{V}\right)$ Common tangent to $C_1$ and $C_3$ is common chord to $C_1$ and $C_3$
$C_1-C_3=0$
$3x+4y+15=0$
Now this line is tangent to parabola
$x^2=8\alpha y$
$x^2=8\alpha \left(\frac{-3x-15}{y}\right)$
$4x^2+24\alpha x+120\alpha =0$
Apply $D=0$ (for tangent, it will have repeated roots)
$\alpha =\frac{10}{3}$

Hence, the correct answer is option: 'A'.