Question

# Let the circles C1: x2+y2=9 and C2: (x−3)2+(y−4)2=16, intersect at the points X and Y. Suppose that another circle C3: (x−h)2+(y−k)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 List−I whose values are given in List−II 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) 2√6(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: C1−C2=0 ∴3x+4y=9 Perpendicular distance from centre of C3 to ZW=∣∣ ∣ ∣ ∣∣3⋅95+4⋅125−9√32+42∣∣ ∣ ∣ ∣∣=65 Length of ZW is, ZW=2√r2−(ZW)2=2 √62−(65)2 =245√6 Perpendicular distance from centre of C1 to XY=∣∣ ∣∣3⋅0+4⋅0−9√32+42∣∣ ∣∣=95 Length of XY=2√32−(95)2=245 ∴Length of ZWLength of XY=√6

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