$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. $A$ straight line $L$,perpendicular to $PT$,is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ $A$ common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ $A$ possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

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:
$(i)$ The centre of $C_3$ is collinear with the centres of $C_1$ and $C_2$.
$(ii)$ $C_1$ and $C_2$ both lie inside $C_3$.
$(iii)$ $C_3$ 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 $List-I$ whose values are given in $List-II$ below:

$List-I$	$List-II$
$(I) \ 2h + k$	$(P) \ 6$
$(II) \ \frac{\text{Length of } ZW}{\text{Length of } XY}$	$(Q) \ \sqrt{6}$
$(III) \ \frac{\text{Area of triangle } MZN}{\text{Area of triangle } ZMW}$	$(R) \ \frac{5}{4}$
$(IV) \ \alpha$	$(S) \ \frac{21}{5}$
	$(T) \ 2\sqrt{6}$
	$(U) \ \frac{10}{3}$

$(1)$ Which of the following is the only $INCORRECT$ combination?
$(1) (IV), (S) \quad (2) (IV), (U) \quad (3) (III), (R) \quad (4) (I), (P)$
$(2)$ Which of the following is the only $CORRECT$ combination?
$(1) (II), (T) \quad (2) (I), (S) \quad (3) (I), (U) \quad (4) (II), (Q)$

If $(\alpha, \beta)$ is a point on the circle whose centre is on the $x$-axis and which touches the line $x + y = 0$ at $(2, -2)$,then the greatest value of $\alpha$ is

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