Answer the following by appropriately matching the lists based on… - Vidyadip

MCQ

Answer the following by appropriately matching the lists based on the information given in the paragraphLet 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 $\text{INCORRECT}$ combination?
$(1) (IV), (S) \ (2) (IV), (U) \ (3) (III), (R) \ (4) (I), (P)$
$(2)$ Which of the following is the only $\text{CORRECT}$ combination?
$(1) (II), (T) \ (2) (I), (S) \ (3) (I), (U) \ (4) (II), (Q)$
Give the answer or quetion ($1$) and ($2$)

✓
$1,4$
B
$1,3$
C
$1,2$
D
$2,4$

✓

Answer

Correct option: A.

$1,4$

$($image$)MC _1+ C _1 C _2+ C _2 N =2 r$
$\Rightarrow 3+5+4=2 r$
$\Rightarrow r =6$
$\Rightarrow$ Radius of $C _3=6$
Suppose centre of $C_3$ be $\left(0+r_4 \cos \theta, 0+r_4 \sin \theta\right),\left\{\begin{array}{l}r_4=C_1 C_3=3 \\ \tan \theta=\frac{4}{3}\end{array}\right\}$
$C _3=\left(\frac{9}{5}, \frac{12}{5}\right)=( h , k )$
$\Rightarrow 2 h + k =6$
Equation of $ZW$ and $XY$ is $3 x +4 y -9=0$
$($common chord of circle $C _1=0$ and $C _2=0)$
$ZW =2 \sqrt{ r ^2- p ^2}=\frac{24 \sqrt{6}}{5} ($where $r =6$ and $p =\frac{6}{5})$
$($image$)$
$XY =2 \sqrt{ r _1^2- p _1^2}=\frac{24}{5} ($where $r _1=3$ and $p _1=\frac{9}{5})$
$($image$)$
$\frac{\text { Length of } ZW }{\text { Length of } XY }=\sqrt{6}$
Let length of perpendicular from $M$ to $ZW$ be $\lambda, \lambda=3+\frac{9}{5}=\frac{24}{5}$
$\frac{\text { Area of } \triangle MZN }{\text { Area of } \triangle ZMW }=\frac{\frac{1}{2}( MN ) \times \frac{1}{2}( ZW )}{\frac{1}{2} \times ZW \times \lambda}=\frac{1}{2} \frac{ MN }{\lambda}=\frac{5}{4}$
$C_3:\left(x-\frac{9}{5}\right)^2+\left(y-\frac{12}{5}\right)^2=6^2$
$C_1: x^2+y^2-9=0$
common tangent to $C _1$ and $C _3$ is common chord of $C _1$ and $C _3$ is $3 x +4 y +15=0$.
Now $3 x+4 y+15=0$ is tangent to parabola $x^2=8 \alpha y$.
$x^2=8 \alpha\left(\frac{-3 x -15}{4}\right)$
$\Rightarrow 4 x ^2+24 \alpha x +120 \alpha=0$
$D =0$
$\Rightarrow \alpha=\frac{10}{3}$

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