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Question

Let the circles C1: x2+y2=9 and C2: (x3)2+(y4)2=16, intersect at the points X and Y. Suppose that another circle C3: (xh)2+(yk)2=r2 satisfies the following conditions:

(i) centre of C3 is collinear with the centres of C1 and C2.

(ii)C1 and C2 both lie inside C3, and

(iii) C3 touches C1 at M and C2 at N

Let the line through X and Y intersect C3 at Z and W, and let a common tangent of C1 and C3 be the tangent to the parabola x2=8αy.
There are some expressions given in ListI whose values are given in ListII below:

List IList II(I)2h+k (P) 6(II)length of ZWlength of XY (Q) 6(III)Area of triangle MZNArea of triangle ZMW (R) 54(IV)α (S) 215(T) 26(U) 103
Which of the following is the only CORRECT combination?

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

The correct option is C (II),(Q)

Centre of C1,C2,C3 are collinear and C1,C2 both lie inside C3.
MN is diameter of C3
MN=MC1+C1C2+C2N
2r=3+32+42+4=12
r=6
where, r is the radius of C3

Suppose centre of C3 be (h,k)=(0+r4cosθ,0+r4sinθ)r4=C1C3=3 and tanθ=43
C3=(95,125)=(h,k)
2h+k=6
Equation of ZW and XY is common chord of circle C1=0 and C2=0


Equation of ZW: C1C2=0
3x+4y=9
Perpendicular distance from centre of C3 to ZW=∣ ∣ ∣ ∣395+4125932+42∣ ∣ ∣ ∣=65
Length of ZW is,
ZW=2r2(ZW)2=2 62(65)2 =2456


Perpendicular distance from centre of C1 to XY=∣ ∣30+40932+42∣ ∣=95
Length of XY=232(95)2=245
Length of ZWLength of XY=6

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