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 $3k = 4h$ …(i) $⇒ MN$ is diameter of ${C}_{3}$ $MN = M{C}_{1} + {C}_{1}{C}_{2} + {C}_{2}N$ \begin{cases}2r = {r}_{1} + {C}_{1}{C}_{2} + {r}_{2} &\left(∵ MN = 2r\right)\end{cases}2r = 3 + \sqrt{{\left(3 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}} + 4r = 6$…(ii)$⇒$Given${C}_{3}$touches${C}_{1}$at$\begin{cases}M & \end{cases}$So$\begin{cases}\left|{C}_{1}{C}_{3}\right| = \left|r - 3\right| & \end{cases}{h}^{2} + {k}^{2} = 9$…(iii) From (i) and (iii)$h = \pm \dfrac{9}{5}$and$k = \pm \dfrac{12}{5}$So centre of${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} = 06x + 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 $ZW = 2\sqrt{{\left(6\right)}^{2} - {\left(\dfrac{6}{5}\right)}^{2}}$ $ZW = \dfrac{24\sqrt{6}}{5}$ $\left(I\right)2h + k = 2 \times \dfrac{9}{5} + \dfrac{12}{5} = \dfrac{30}{5} = 6$ $\left(II\right)\dfrac{LengthofZW}{LengthofXY} = \sqrt{6}$ $\left(III\right)\dfrac{Areaof ∆ MZN}{Areaof ∆ ZMW} = \dfrac{\dfrac{1}{2} \times MN \times PZ}{\dfrac{1}{2} \times ZW \times MP}$ $=$\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}$=\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}$$= \dfrac{5}{4} \left(IV\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\alphay {x}^{2} = 8\alpha\left(\dfrac{- 3x - 15}{y}\right)$4{x}^{2} + 24\alphax + 120\alpha = 0$Apply$D = 0$(for tangent, it will have repeated roots)$\alpha = \dfrac{10}{3}$