The equation of a common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ 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:

$A$ circle with centre $P$ is tangent to the negative $x$-axis and negative $y$-axis and is externally tangent to a circle with centre $(-6, 0)$ and radius $2$. What is the sum of all possible radii of the circle with centre $P$?

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 area of the triangle formed by joining the origin to the points of intersection of the line $x\sqrt{5} + 2y = 3\sqrt{5}$ and the circle $x^2 + y^2 = 10$ is

The circle ${x^2} + {y^2} = 4$ cuts the line joining the points $A(1, 0)$ and $B(3, 4)$ in two points $P$ and $Q$. Let $\frac{BP}{PA} = \alpha$ and $\frac{BQ}{QA} = \beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation