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: $\left(i\right) \quad$ centre of $C_{3}$ is collinear with the centres of $C_{1}$ and $C_{2}$, $\left(ii\right) \quad C_{1}$ and $C_{2}$ both lie inside $C_{3}$, and $\left(iii\right)\quad C_{3}$ 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: Which of the following is the only CORRECT combination?

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

Answer 1

The Correct Option is C

$MC_{1}+C_{1}C_{2}+C_{2}N=2r$ $\Rightarrow 3+5+4=2r \Rightarrow r=6 \Rightarrow$ Radius of $C_{3}=6$ Suppose centre of $C_{3} be \left(0+r_{4}\,cos\,\theta, \theta+r_{4}\,sin\,\theta\right),\begin{Bmatrix}r_{4}=C_{1}&C_{3}=3\\ tan&\theta=\frac{4}{3}\end{Bmatrix}$ $C_{3}=\left(\frac{9}{5}, \frac{12}{5}\right)=\left(h.k\right) \Rightarrow 2h+k=6$ Equation of $ZW$ and $XY is 3x + 4y - 9 = 0$ $\left(common\, chord\, of\, circle\, C_{1} = 0\, and \,C_{2} = 0\right)$ $ZW=2\sqrt{r^{2}-p^{2}}=\frac{24\sqrt{6}}{5}\left(where\,r=6\,and\,p=\frac{6}{5}\right)$ $XY=2\sqrt{r^{2}_{1}=p^{2}_{1}}=\frac{24}{5}\left(where\,r_{1}=3\,and\,p_{1}=\frac{9}{5}\right)$ $\frac{Length\, of ZW}{Length\, of XY}=\sqrt{6}$ Let length of perpendicular from $M$ to $ZW be \lambda, \lambda=3+\frac{9}{5}=\frac{24}{5}$ $\frac{Area\,of \, \Delta MZN}{Area\, of \,\Delta ZMW}=\frac{\frac{1}{2}\left(MN\right)\times\frac{1}{2}\left(ZW\right)}{\frac{1}{2}\times ZW\times\lambda}=\frac{1}{2} \frac{MN}{\lambda}=\frac{5}{4}$ $C_{3} : \left(x-\frac{9}{5}\right)^{2}+\left(y-\frac{12}{5}\right)^{2}=6^{2}$ $C_{1} : x^{2}+y^{2}-9=0$ 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 $x^2 = 8\alpha y$. $x^{2}=8\alpha\left(\frac{-3x-15}{4}\right) \Rightarrow 4x^{2}+24\alpha x+120\alpha=0$ $D=0 \Rightarrow \alpha=\frac{10}{3}$