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JEE Advanced 2019 Paper 2

Section: Mathematics

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Question : 53 of 54

Marks: +1, -0

Answer the following by appropriately matching the lists based on the information given in the paragraph.

Let the circles

C_{1}:x^{2}+y^{2}=9

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 :

C_3

is collinear with the centres of

C_1

C_2

C_1

C_2

both lie inside

C_3

C_3

C_1

C_2

Let the line through X and Y intersect

C_3

at Z and W, and let a common tangent of

C_1

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

List I	List II
I) $2h+k$	P) 6
II) $\frac{\;\text{Length of }\;ZW}{\;\text{Length of }\;XY}$	Q) $\sqrt{6}$
III) $\frac{\;\text{Area of triangle MZN}\;}{\;\text{Area of triangle ZMW}\;}$	R) $\frac{5}{4}$
IV) α	S) $\frac{21}{5}$
	T) $2\sqrt{6}$
	U) $\frac{10}{3}$

Which of the following is the only CORRECT combination?

[JEE Adv 2019 P2]

(I), (U)
(II), (T)
(II), (Q)
(I), (S)

Solution:

C_{1}=x^{2}+y^{2}=9\Rightarrow O(0,0)

C_{2}=(x-3)^{2}+(y-4)^{2}=16\Rightarrow x^{2}+y^{2}-6x-8y+9=0\Rightarrow C_{2}(3,4)

C_{3}=(x-h)^{2}+(y-k)^{2}=r^{2}

Let the centre of

\underline{C}_3

be C.

\therefore OC:CC_{2}=3:2\Rightarrow C=\left(\; \frac{9}{5},\; \frac{12}{5}\right)

\Rightarrow 2h+k=6, 2r=3+5+4=12\Rightarrow r=6

Eq. of common chord of

C_1

and

C_2

is

3x+4y-9=0

. Let P be the point of intersection of

OC_1

and common chord.

\therefore CP=\; \frac{6}{5}, ZP=WP=\; \frac{12\sqrt{6}}{5}, MP=\; \frac{24}{5}

XY=2\sqrt{9-\; \frac{81}{25}}=\; \frac{24}{5}

\therefore\; \frac{ZW}{XY}=\sqrt{6}

\; \frac{[\triangle MZN]}{\triangle ZMW}=\; \frac{5}{4}

Equation of the common chord of

C_1

and

C_3

is

3x+4y+15=0\Rightarrow \alpha=\; \frac{10}{3}

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