Given circles

$C_1: x^2+y^2=9$ has centre $O_1=(0,0)$ and radius $r_1=3$.

$C_2:(x-3)^2+(y-4)^2=16$ has centre $O_2=(3,4)$ and radius $r_2=4$.

Distance between centres: $O_1O_2=\sqrt{3^2+4^2}=5$

Find circle $C_3$

Let centre of $C_3$ be $(h,k)$ and radius $r$.

Since $C_3$ contains both circles internally and touches them, its centre lies on the line joining $O_1$ and $O_2$.

Also, for internal tangency: $CO_1=r-3, \qquad CO_2=r-4$ where $C$ is centre of $C_3$.

Because $C$ lies on line $O_1O_2$ and outside the segment on the side of $O_2$, we have $CO_1-CO_2= (r-3)-(r-4)=1$

On the line through $(0,0)$ and $(3,4)$, take $C=(3,4)+t(3,4)=(3+3t,4+4t)$ Then $CO_2=5t, \qquad CO_1=5(1+t)$ and indeed $CO_1-CO_2=5$ if $C$ is beyond $O_2$, so better use tangency equations directly.

Let $CO_1=d_1$, $CO_2=d_2$. Since $C,O_1,O_2$ are collinear and $C$ is beyond $O_2$, $d_1=d_2+5$ Also, $d_1=r-3,\quad d_2=r-4$ Hence these are consistent automatically: $r-3=(r-4)+5$ which gives $1=1$.

So there are infinitely many such circles unless we interpret the circle touching both externally from outside while containing both inside. Then centre must lie beyond $O_2$ and from geometry, $d_1=r-3,\ d_2=r-4,\ d_1-d_2=5$ But then $(r-3)-(r-4)=1\neq 5$ impossible.

Therefore centre must lie between $O_1$ and $O_2$? Let us test: $d_1+d_2=5$ with $d_1=r-3,\ d_2=r-4$ so $(r-3)+(r-4)=5 \Rightarrow 2r=12 \Rightarrow r=6$ Then $d_1=3,\ d_2=2$ So centre divides $O_1O_2$ internally in ratio $3:2$.

Hence $C=\left(\frac{3}{5}\cdot 3,\frac{3}{5}\cdot 4\right)=\left(\frac95,\frac{12}5\right)$ and $C_3:\left(x-\frac95\right)^2+\left(y-\frac{12}5\right)^2=36$

Radical axis of $C_1$ and $C_2$

Subtract equations: $x^2+y^2-[(x-3)^2+(y-4)^2]=9-16=-7$ $x^2+y^2-(x^2-6x+9+y^2-8y+16)=-7$ $6x+8y-25=-7$ $6x+8y=18$ $3x+4y=9$

So line $XY$ is $3x+4y=9$

Points where this line cuts $C_3$

Since centre $C=(9/5,12/5)$ lies on $3x+4y=15$, the perpendicular distance from $C$ to line $3x+4y=9$ is $\frac{|15-9|}{\sqrt{3^2+4^2}}=\frac65$

Chord length in $C_3$ cut by this line:

Common tangent of $C_1$ and $C_3$

Since $C_1$ and $C_3$ touch internally at $M$, their common tangent is the tangent at $M$.

Point $M$ lies on line joining centres. Since $O_1C=3$ and radius of $C_1$ is $3$, $M=(0,0)+\frac{3}{5}(3,4)=\left(\frac95,\frac{12}5\right)$ So indeed $M$ coincides with centre of $C_3$? That is impossible, so let us correct.

Since $CO_1=3$ and radius of $C_1=3$, touching point $M$ is midpoint of $O_1C$: $M=\left(\frac{9}{10},\frac65\right)$

Tangent to $C_1: x^2+y^2=9$ at $\left(\frac{9}{5}?\right)$ is not possible because point must lie on $C_1$.

Instead, direction of line $O_1C$ is $(3,4)$, so touching point on $C_1$ is $M=\left(\frac{9}{5},\frac{12}{5}\right)$ and indeed $\left(\frac95\right)^2+\left(\frac{12}5\right)^2=\frac{81+144}{25}=9$ so $M$ is on $C_1$.

Tangent at $M$ to $C_1$ is $xx_1+yy_1=9$ $\frac95x+\frac{12}5y=9$ $3x+4y=15$

Parabola tangent condition

For parabola $x^2=8\alpha y$ we compare with standard form $x^2=4ay$, so $4a=8\alpha\Rightarrow a=2\alpha$.

A tangent to $x^2=4ay$ with slope $m$ is $y=mx-am^2$

Given line $3x+4y=15 \Rightarrow y=-\frac34x+\frac{15}{4}$ so slope $m=-\frac34$ and intercept $\frac{15}{4}$.

Therefore $-am^2=\frac{15}{4}$ Since $m=-3/4$, $-a\cdot \frac{9}{16}=\frac{15}{4}$ giving $a<0$, impossible for usual upward parabola, but mathematically $a=-\frac{20}{3}$ Hence $2\alpha=a=-\frac{20}{3} \Rightarrow \alpha=-\frac{10}{3}$

Issue with the question data

The problem asks to match List-I and List-II, but the actual contents of List-I items (I), (II), (III), (IV) and List-II values (P), (Q), (R), (S), (U) are not provided in the prompt.

Therefore, it is impossible to verify which combination is incorrect from the options A, B, C, D.

Comparison with stored answer

Since the essential matching lists are missing, I cannot logically derive option $B$ (or any option) from the given information alone.

Hence I must disagree with the stored answer due to incomplete question statement.