JEE Advanced 2019 · Paper 2 · Mathematics Q18

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MathematicsList match+3 / -1Question 54 of 54

Passage

Answer Q.17 and Q.18 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 expressions given in the List-I whose values are given in List-II below:

Which of the following is the only INCORRECT combination?

List-I

I
$2h + k$
II
$\dfrac{\text{Length of }ZW}{\text{Length of }XY}$
III
$\dfrac{\text{Area of triangle }MZN}{\text{Area of triangle }ZMW}$
IV
$\alpha$

List-II

P
6
Q
$\sqrt6$
R
$\dfrac54$
S
$\dfrac{21}5$
T
$2\sqrt6$
U
$\dfrac{10}3$

(I), (P)

(IV), (U)

(III), (R)

(IV), (S)

Answer and worked solution

Official answer: D

(III) Ratio of areas. Z is $\frac{12\sqrt6}5$ from the line MN, and M is $\frac{24}5$ from the chord ZW. So
$\frac{[MZN]}{[ZMW]}=\frac{\frac12\cdot12\cdot\frac{12\sqrt6}5}{\frac12\cdot\frac{24\sqrt6}5\cdot\frac{24}5}=\frac54,$
(III) → (R).

(IV) α. The common tangent of $C_1$ and $C_3$ is at M, perpendicular to OM: $3x+4y+15=0$ . Substituting into $x^2=8\alpha y$ :
$x^2+6\alpha x+30\alpha=0.$
A tangent needs a zero discriminant, $36\alpha^2=120\alpha$ , so $\alpha=\frac{10}3$ : (IV) → (U).

The options. (I) → (P), (IV) → (U) and (III) → (R) are all correct. (IV) → (S) is the only incorrect pairing.

(D)

Topics:Tangents and normals Parabola Family of circles

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