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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 INCORRECT combination?

A
(I),(P)
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B
(IV),(U)
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C
(III),(R)
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D
(IV),(S)
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Solution

The correct option is D (IV),(S)

Let length of perpendicular from M to ZW be λ
λ=3+95=245
Area of triangle MZNArea of triangle ZMW=12(MN)12(ZW)12(ZW)λ=54
C3: (x95)2+(y125)2=62
C1: x2+y2=9
Common tangent to C1 and C3 is C1C3=03x+4y+15=0
Now,3x+4y+15=0 is tangent to parabola
x2=8αyx2=8α(3x154)
x2+6αx+30α=0
D=0=b24ac
α=103
So, α=215 is an incorrect combination.

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