Let the circles C _ 1 : x ^ 2 + y ^ 2 = 9 and C _ 2 : (x - 3) ^ 2 +...

Given centre of ${C}_{1}, {C}_{2}$ and ${C}_{3}$ are collinear hence
\left|$\begin{cases}0 & 0 & 1 \\ 3 & 4 & 1 \\ h & k & 1\end{cases}$\right| = 0$<br/>$3k = 4h$…(i)<br/>$⇒ MN$is diameter of${C}_{3}$<br/>$MN = M{C}_{1} + {C}_{1}{C}_{2} + {C}_{2}N$<br/>$\begin{cases}2r = {r}_{1} + {C}_{1}{C}_{2} + {r}_{2} & \left(∵ MN = 2r\right)\end{cases}$<br/>$2r = 3 + \sqrt{{\left(3 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}} + 4$<br/>$r = 6$…(ii)<br/>$⇒$Given${C}_{3}$touches${C}_{1}$at$\begin{cases}M & \end{cases}$<br/>So$\begin{cases}\left|{C}_{1}{C}_{3}\right| = \left|r - 3\right| & \end{cases}$<br/>${h}^{2} + {k}^{2} = 9$…(iii)<br/>From (i) and (iii)<br/>$h = \pm \dfrac{9}{5}$and$k = \pm \dfrac{12}{5}$<br/>So centre of${C}_{3}$will be \left(\dfrac{9}{5}, \dfrac{12}{5}\right)$
$⇒$ 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
${C}_{1}P = \dfrac{9}{5}$
$⇒$ now in $∆ {C}_{1}Py$
${\left({C}_{1}P\right)}^{2} + {\left(PY\right)}^{2} = {\left({C}_{1}Y\right)}^{2}$
$\begin{cases}\dfrac{81}{25} + {\left(PY\right)}^{2} = \left(9\right) & \left(∵ {C}_{1}Y = {r}_{1} = 3\right)\end{cases}$
${\left(PY\right)}^{2} = \dfrac{144}{25} ⇒ PY = \dfrac{12}{5}$
Length of $XY = 2PY = \dfrac{24}{5}$
$⇒$ Line $ZW$ is line $XY$
Equation of $ZW = 3x + 4y = 9$
Distance of ${C}_{3}$ from $ZW = \left|\dfrac{3\left(\dfrac{9}{5}\right) + 4\left(\dfrac{12}{5}\right) - 9}{5}\right|$
$ZW = \dfrac{6}{5}$
now $ZW = 2\sqrt{{\left(6\right)}^{2} - {\left(\dfrac{6}{5}\right)}^{2}}$
$ZW = \dfrac{24\sqrt{6}}{5}$

$\left(I\right)2h + k = 2 \times \dfrac{9}{5} + \dfrac{12}{5} = \dfrac{30}{5} = 6$
\left(II\right)\dfrac{LengthofZW}{LengthofXY} = \sqrt{6}$<br/>\left(III\right)\dfrac{Areaof ∆ MZN}{Areaof ∆ ZMW} = \dfrac{\dfrac{1}{2} \times MN \times PZ}{\dfrac{1}{2} \times ZW \times MP}$
$=$\begin{cases}\dfrac{\dfrac{1}{2}MN\left(\dfrac{1}{2}ZW\right)}{\dfrac{1}{2} \times ZW\left(MG + GP\right)} & \left(∴ PZ = \dfrac{1}{2}ZW\right)\end{cases}<br/>$=\begin{array}{ll}\dfrac{\dfrac{1}{2} \times 12 \times \dfrac{12\sqrt{6}}{5}}{\dfrac{1}{2} \times \dfrac{24\sqrt{8}}{5}\left(\dfrac{24}{5}\right)} & ∴ M{C}_{1} = 3 \\ ∴ {C}_{1}A = \dfrac{9}{5}\end{array}$$$<br/>$= \dfrac{5}{4}$<br/>$\left(IV\right)$Common tangent to${C}_{1}$and${C}_{3}$is common chord to${C}_{1}$and${C}_{3}$<br/>${C}_{1} - {C}_{3} = 0$<br/>$3x + 4y + 15 = 0$<br/>Now this line is tangent to parabola<br/>${x}^{2} = 8\alphay$<br/>${x}^{2} = 8\alpha\left(\dfrac{- 3x - 15}{y}\right)$<br/>$4{x}^{2} + 24\alphax + 120\alpha = 0$<br/>Apply$D = 0$(for tangent, it will have repeated roots)<br/>$\alpha = \dfrac{10}{3}