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Yukti Vault

Circle - JEE Mathematics question (JEE Advanced 2019 Paper 2)

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

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 expressions given in the List-I whose values are given in List-II below: $$\begin{array}{ll}\text{List-I} & \text{List-II}\\ \text{(I)}\ 2h+k & \text{(P)}\ 6\\ \text{(II)}\ \dfrac{\text{Length of } ZW}{\text{Length of } XY} & \text{(Q)}\ \sqrt6\\ \text{(III)}\ \dfrac{\text{Area of triangle } MZN}{\text{Area of triangle } ZMW} & \text{(R)}\ \dfrac54\\ \text{(IV)}\ \alpha & \text{(S)}\ \dfrac{21}{5}\\ & \text{(T)}\ 2\sqrt6\\ & \text{(U)}\ \dfrac{10}{3}\end{array}$$
JEE Advanced 2019 Paper 2Question 1
Which of the following is the only CORRECT combination?
Circle
Options for question 1, choose one

Solution

Concept. Data (from the paragraph of this question): $C_1: x^2+y^2=9$, $C_2:(x-3)^2+(y-4)^2=16$ meet at $X, Y$. $C_3:(x-h)^2+(y-k)^2=r^2$ touches $C_1$ at $M$ and $C_2$ at $N$ internally, with $M, N$ on the line of centres and diameter $=$ sum of radii and distance between centres. The radical axis $XY$ cuts $C_3$ at $Z, W$; the tangent at $M$ also touches $x^2=8\alpha y$. Lists: (I) $2h+k$, (II) $\frac{ZW}{XY}$, (III) $\frac{\text{ar}(MZN)}{\text{ar}(ZMW)}$, (IV) $\alpha$; (P) $6$, (Q) $\sqrt6$, (R) $\frac54$, (S) $\frac{21}{5}$, (T) $2\sqrt6$, (U) $\frac{10}{3}$. Formulas used. (1) Half-chord $=\sqrt{r^2-d^2}$, $d$ = distance of the chord from the centre.(2) Radical axis of $S_1=0, S_2=0$: $S_1-S_2=0$.(3) Tangent to $x^2+y^2=a^2$ at $(x_1,y_1)$: $xx_1+yy_1=a^2$.Step 1 - $C_3$ and its centre. $O_1=(0,0), r_1=3$; $O_2=(3,4), r_2=4$; $O_1O_2=5$. Diameter of $C_3$ $=3+5+4=12$, so $r=6$. Unit vector $O_1\to O_2$ is $\left(\frac35,\frac45\right)$: $$M=-3\left(\tfrac35,\tfrac45\right)=\left(-\tfrac95,-\tfrac{12}5\right),\quad N=(3,4)+4\left(\tfrac35,\tfrac45\right)=\left(\tfrac{27}5,\tfrac{36}5\right).$$ Centre $=$ midpoint of $MN=\left(\frac95,\frac{12}5\right)$, so $2h+k=\frac{18}5+\frac{12}5=6$. (I) $\to$ (P). Step 2 - Radical axis, use (2). $(x^2+y^2-9)-(x^2+y^2-6x-8y+9)=0\Rightarrow 3x+4y=9$. Step 3 - $XY$ and $ZW$, use (1). Distance from $O_1$: $\frac95$, so $XY=2\sqrt{9-\frac{81}{25}}=\frac{24}5$. Distance from $O_3$: $\frac{|27/5+48/5-9|}{5}=\frac65$, so $ZW=2\sqrt{36-\frac{36}{25}}=\frac{24\sqrt6}{5}$. Hence $\frac{ZW}{XY}=\sqrt6$. (II) $\to$ (Q). Step 4 - Area ratio. $ZW\perp MN$ (both chords are perpendicular to the line of centres). Let $F$ be the foot from $M$ on $ZW$. $$\text{ar}(MZN)=\tfrac12\cdot MN\cdot\tfrac{ZW}{2},\qquad \text{ar}(ZMW)=\tfrac12\cdot ZW\cdot MF.$$ Ratio $=\frac{MN}{2MF}$. $MF=\frac{|-27/5-48/5-9|}{5}=\frac{24}5$, so ratio $=\frac{12}{48/5}=\frac54$. (III) $\to$ (R). Step 5 - $\alpha$, use (3). Tangent to $C_1$ at $M$: $-\frac95x-\frac{12}5y=9\Rightarrow 3x+4y+15=0$. Put $y=-\frac{3x+15}{4}$ in $x^2=8\alpha y$: $x^2+6\alpha x+30\alpha=0$. Tangency: $36\alpha^2-120\alpha=0\Rightarrow\alpha=\frac{10}3$. (IV) $\to$ (U). Step 6 - Check options. Only (II),(Q) is a correct pair. Answer: C

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