The angle at which the circles $(x - 1)^2 + y^2 = 10$ and $x^2 + (y - 2)^2 = 5$ intersect is

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?
$(1) (II), (T) \quad (2) (I), (S) \quad (3) (I), (U) \quad (4) (II), (Q)$

The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is:

$B$ and $C$ are fixed points having coordinates $(3, 0)$ and $(-3, 0)$ respectively. If the vertical angle $\angle BAC$ is $90^o$,then the locus of the centroid of the $\Delta ABC$ has the equation:

For $n \in N$,let $S_{n} = \{ z \in C : |z - 3 + 2i| = \frac{n}{4} \}$ and $T_{n} = \{ z \in C : |z - 2 + 3i| = \frac{1}{n} \}$. Then the number of elements in the set ${ n \in N : S_{n} \cap T_{n} = \phi }$ is.

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then
$(A)$ $SP=2\sqrt{5}$
$(B)$ $SQ:QP=(\sqrt{5}+1):2$
$(C)$ the $x$-intercept of the normal to the parabola at $P$ is $6$
$(D)$ the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$