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Question:

Let the circle 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 condition. (i) The center of C3 is collinear with the centers of C1 and C2. (ii) C1 and C2 both lie inside C3. (iii) C3 touches C1 at M and N. Let the line through X and Y intersect C3 at Z and W. Let a common tangent of C1 and C3 be a tangent to the parabola x² = 8αy. There are some expressions given in the list - I whose values are given in list II below:

Sr. No. List I Sr. No. List II
  1.  | 2h + k   | P.       | 6
    
  2.  | length of ZW / length of XY | Q.       | 6
    
  3.  | Area of ΔMZN / Area of ΔZMW | R.       | 544
    
  4.  | α        | S.       | 215
    

T. | 26
U. | 103

Which of the following is the only correct combination?

(I) - (S)

(II) - (T)

(II) - (Q)

(I) - (U)

Solution:

Correct option is C (II) - (Q)

I. 2r = MN = 3 + √(3² + 4² + 4) = 12 ⇒ r = 6
Center C of circle C3 lies on y = (4/3)x
Let C(h, 4/3h)
OC = MC - OM = √(12²/2) - 3 = 3
∴ h² + (16/9)h² = 9 ⇒ 25h²/9 = 9 ⇒ h = 9/5
k = (4/3)h = 12/5
∴ 2h + k = 18/5 + 12/5 = 6

II. Equation of line ZW
C1 = C2 ⇒ 3x + 4y = 9
Length of XY = 2√(3² - (9/5)²) = 24/5
Distance of ZW from C |(3 × 9/5 + 4 × 12/5 - 9)|/√(3² + 4²) = 6/5
Length of ZW = 2√(6² - (6/5)²) = 24√6/5
∴ length of ZW / length of XY = 6

III. Area of ΔMZN = (1/2).NM.(1/2 ZW) = 72/65
Area of ΔZMW = (1/2).ZW.(OM + OP) = (1/2)(24√6/5)(3 + 9/5) = 288√6/25
∴ Area of ΔMZN / Area of ΔZMW = 54

IV. Slope of tangent to C1 at M = -1/(4/3) = -3/4
∴ Equation of tangent y = (-3/4)x - 3√(1 + 9/16) = (-3/4)x - 15/4
⇒ 3x + 4y = -15 (i)
Tangent to x² = 4(2α)y is x = m'y + 2αm' (ii)
Compare (i) and (ii)
m' = -4/3 and 2αm' = -15
⇒ α = 103