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← JEE Advanced 2019 · Paper 2
MathematicsList match+3 / -1Question 53 of 54

JEE Advanced 2019 · Paper 2, Mathematics question 17

Passage

Answer Q.17 and Q.18 by appropriately matching the lists based on the information given in the paragraph.

Let the circles C1:x2+y2=9C_1: x^2 + y^2 = 9 and C2:(x−3)2+(y−4)2=16C_2: (x-3)^2 + (y-4)^2 = 16, intersect at the points XX and YY. Suppose that another circle C3:(x−h)2+(y−k)2=r2C_3: (x-h)^2 + (y-k)^2 = r^2 satisfies the following conditions:

(i) centre of C3C_3 is collinear with the centres of C1C_1 and C2C_2,

(ii) C1C_1 and C2C_2 both lie inside C3C_3, and

(iii) C3C_3 touches C1C_1 at MM and C2C_2 at NN.

Let the line through XX and YY intersect C3C_3 at ZZ and WW, and let a common tangent of C1C_1 and C3C_3 be a tangent to the parabola x2=8αyx^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 CORRECT combination?

List-I

  • I
    2h+k2h + k
  • II
    Length of ZWLength of XY\dfrac{\text{Length of }ZW}{\text{Length of }XY}
  • III
    Area of triangle MZNArea of triangle ZMW\dfrac{\text{Area of triangle }MZN}{\text{Area of triangle }ZMW}
  • IV
    α\alpha

List-II

  • P
    6
  • Q
    6\sqrt6
  • R
    54\dfrac54
  • S
    215\dfrac{21}5
  • T
    262\sqrt6
  • U
    103\dfrac{10}3
Choose one option
Answer and worked solution

Official answer: C

The circles. C1C_1 has centre O and radius 3; C2C_2 has centre (3,4)(3,4) and radius 4; the centres are 5 apart.

C3C_3. It touches both internally along the line of centres:

  • M=(−95,−125)M=\left(-\frac95,-\frac{12}5\right) and N=(275,365)N=\left(\frac{27}5,\frac{36}5\right);
  • its diameter is 3+5+4=123+5+4=12, so its radius is 6 and its centre is (95,125)\left(\frac95,\frac{12}5\right).

So 2h+k=62h+k=6, (I) → (P).

Chords. The common chord XY is 3x+4y=93x+4y=9.

  • Its distance from O is 95\frac95, so XY=29−8125=245XY=2\sqrt{9-\frac{81}{25}}=\frac{24}5.
  • Its distance from C3C_3's centre is 65\frac65, so ZW=236−3625=2465ZW=2\sqrt{36-\frac{36}{25}}=\frac{24\sqrt6}5.

So ZWXY=6\dfrac{ZW}{XY}=\sqrt6, (II) → (Q).

(C)