The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.
$1.$ Equation of a common tangent with positive slope to the circle as well as to the hyperbola is:
$(A) 2x-\sqrt{5}y-20=0$
$(B) 2x-\sqrt{5}y+4=0$
$(C) 3x-4y+8=0$
$(D) 4x-3y+4=0$
$2.$ Equation of the circle with $AB$ as its diameter is:
$(A) x^2+y^2-12x+24=0$
$(B) x^2+y^2+12x+24=0$
$(C) x^2+y^2+24x-12=0$
$(D) x^2+y^2-24x-12=0$

If the lines $kx + 2y - 4 = 0$ and $5x - 2y - 4 = 0$ are conjugate with respect to the circle $x^2 + y^2 - 2x - 2y + 1 = 0$,then $k$ is equal to

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter through $A$ is

$A$ parabola $y = ax^2 + bx + c$ crosses the $x$-axis at $(\alpha, 0)$ and $(\beta, 0)$,both to the right of the origin. $A$ circle also passes through these two points. The length of a tangent from the origin to the circle is:

$2x - 3y + 1 = 0$ and $4x - 5y - 1 = 0$ are the equations of two diameters of the circle $S \equiv x^2 + y^2 + 2gx + 2fy - 11 = 0$. $Q$ and $R$ are the points of contact of the tangents drawn from the point $P(-2, -2)$ to this circle. If $C$ is the centre of the circle $S = 0$,then the area (in square units) of the quadrilateral $PQCR$ 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?
$(1) (II), (T) \quad (2) (I), (S) \quad (3) (I), (U) \quad (4) (II), (Q)$