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 :
$(i)$ centre of $C _3$ is collinear with the centres of $C _1$ and $C _2$
$(ii)$ $C _1$ and $C _2$ both lie inside $C _3$, and
$(iii)$ $C _3$ 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 expression given in the $List-I$ whose values are given in $List-II$ below:
$List-I$ | $List-II$ |
$(I)$ $2 h + k$ | $(P)$ $6$ |
$(II)$ $\frac{\text { Length of } ZW }{\text { Length of } XY }$ | $(Q)$ $\sqrt{6}$ |
$(III)$ $\frac{\text { Area of triangle } MZN }{\text { Area of triangle ZMW }}$ | $(R)$ $\frac{5}{4}$ |
$(IV)$ $\alpha$ | $(S)$ $\frac{21}{5}$ |
$(T)$ $2 \sqrt{6}$ | |
$(U)$ $\frac{10}{3}$ |
($1$) Which of the following is the only INCORRECT combination?
$(1) (IV), (S)$ $(2) (IV), (U)$ $(3) (III), (R)$ $(4) (I), (P)$
($2$) Which of the following is the only CORRECT combination?
$(1) (II), (T)$ $(2) (I), (S)$ $(3) (I), (U)$ $(4) (II), (Q)$
Give the answer or quetion ($1$) and ($2$)
$1,4$
$1,3$
$1,2$
$2,4$
Suppose $S_1$ and $S_2$ are two unequal circles, $A B$ and $C D$ are the direct common tangents to these circles. A transverse common tangent $P Q$ cuts $A B$ in $R$ and $C D$ in $S$. If $A B=10$, then $R S$ is
The value of $'c'$ for which the set, $\{(x, y) | x^2 + y^2 + 2x \le 1 \} \cap \{(x, y) | x - y + c \ge 0\}$ contains only one point in common is :
The circles $x^2 + y^2 + 2x -2y + 1 = 0$ and $x^2 + y^2 -2x -2y + 1 = 0$ touch each other :-
If the circles of same radius a and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally, then $a =$
If the circles $x^{2}+y^{2}+6 x+8 y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y$ $= k +6 \sqrt{3}+8 \sqrt{6}, k >0$, touch internally at the point $P(\alpha, \beta)$, then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$