Question

Let the circles $C_1:x_2+y_2=9$ and $C_2:{\left(x—3\right)}^2+{\left(y—4\right)}^2=16$, intersect at the points
X and Y. Suppose that another circle $C_3:{\left(x—h\right)}^2+{\left(y—k\right)}^2=r^2$ satisfies the following conditions:
$\left(i\right)$ centre of $C_3$ is collinear with the centres of $C_1$ and $C_2$,
$\left(ii\right)C_1$ and $C_2$ both lie inside $C_3$, and
$\left(iii\right)C_3$ 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 $x^2=8ay.$
There are some expressions given in the List-I whose values are given in List-II below:

List-I	List-II
(I) 2 h + k	(p) 6
(II) $\frac{LengthofZW}{LengthofXY}$	(Q) $\sqrt{6}$
(III) $\frac{AreaoftriangleMZN}{AreaoftriangleZMW}$	(R) $\frac{5}{4}$
(IV) $\alpha$	(S) $\frac{21}{5}$
	(T) $2\sqrt{6}$
	(U) $\frac{10}{3}$

Question : Which of the following is the only CORRECT combination?

• A. (I), (S)
• B. (I),(U)
• C. (II), (Q)
• D. (II), (T)

Answer 1

Answer: (C)

$M{C}_1+C_1{C}_2+C_{2}N=2r$
$\Rightarrow 3+5+4=2r\Rightarrow r=6\Rightarrow$ Radius of $C_3=6$
Suppose centre of $C_{3}be\left(0+r_{4}cos\theta ,\theta +r_{4}sin\theta \right),\left\{\begin{array}{cc}{r}_4=C_1& C_3=3\\ tan& \theta =\frac{4}{3}\end{array}\right\}$
$C_3=\left(\frac{9}{5},\frac{12}{5}\right)=\left(h.k\right)\Rightarrow 2h+k=6$
Equation of $ZW$ and $XYis3x+4y-9=0$
$\left(commonchordofcircle{C}_1=0and{C}_2=0\right)$
$ZW=2\sqrt{r^2-p^2}=\frac{24\sqrt{6}}{5}\left(wherer=6andp=\frac{6}{5}\right)$
$XY=2\sqrt{r_1^2=p_1^2}=\frac{24}{5}\left(where{r}_1=3and{p}_1=\frac{9}{5}\right)$
$\frac{LengthofZW}{LengthofXY}=\sqrt{6}$
Let length of perpendicular from $M$ to $ZWbe\lambda ,\lambda =3+\frac{9}{5}=\frac{24}{5}$
$\frac{Areaof\Delta MZN}{Areaof\Delta ZMW}=\frac{\frac{1}{2}\left(MN\right)\times \frac{1}{2}\left(ZW\right)}{\frac{1}{2}\times ZW\times \lambda }=\frac{1}{2}\frac{MN}{\lambda }=\frac{5}{4}$
$C_3:{\left(x-\frac{9}{5}\right)}^2+{\left(y-\frac{12}{5}\right)}^2=6^2$
$C_1:x^2+y^2-9=0$
common tangent to $C_1$ and $C_3$ is common chord of $C_1$ and $C_3$ is $3x+4y+15=0.$
Now $3x+4y+15=0$ is tangent to parabola $x^2=8\alpha y$.
$x^2=8\alpha \left(\frac{-3x-15}{4}\right)\Rightarrow 4x^2+24\alpha x+120\alpha =0$
$D=0\Rightarrow \alpha =\frac{10}{3}$