JEE Challenger
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MathHardAdv 2019 • Paper 2

Matching Circle Intersection and Collinear Centers Properties

Comprehension Passage

Answer Q.17 and Q.18 by appropriately matching the lists based on the information given below.

Consider the circles C1:x2+y2=9C_1 : x^2 + y^2 = 9 and C2:(x−3)2+(y−4)2=16C_2 : (x - 3)^2 + (y - 4)^2 = 16, which intersect at two distinct points XX and YY. Let another circle C3:(x−h)2+(y−k)2=r2C_3 : (x - h)^2 + (y - k)^2 = r^2 satisfy the following: (i) the centre of C3C_3 is collinear with the centres of C1C_1 and C2C_2, (ii) circles C1C_1 and C2C_2 both lie inside C3C_3, and (iii) C3C_3 touches C1C_1 at MM and C2C_2 at NN. Let the line passing through XX and YY intersect C3C_3 at points ZZ and WW, and let a common tangent to C1C_1 and C3C_3 be a tangent to the parabola x2=8αyx^2 = 8\alpha y. Some expressions are given in List-I whose values are listed in List-II:

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\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{(I) } 2h + k & \text{(P) } 6 \\ \text{(II) } \frac{\text{Length of } ZW}{\text{Length of } XY} & \text{(Q) } \sqrt{6} \\ \text{(III) } \frac{\text{Area of triangle } MZN}{\text{Area of triangle } ZMW} & \text{(R) } \frac{5}{4} \\ \text{(IV) } \alpha & \text{(S) } \frac{21}{5} \\ & \text{(T) } 2\sqrt{6} \\ & \text{(U) } \frac{10}{3} \end{array}

Which of the following is the only CORRECT combination?

Options

A

(I), (S)

B

(I), (U)

C

(II), (Q)

Correct
D

(II), (T)

Topics & Concepts

Conic SectionsCirclesParabolaTangents and Normals

Step-by-Step Solution

To find the correct combination, we evaluate the expressions step by step.

Step 1: Determine the properties of circles C1,C2,C_1, C_2, and C3C_3

  1. Circle C1C_1: Equation: x2+y2=9\text{Equation: } x^2 + y^2 = 9

    • Centre: O1=(0,0)O_1 = (0, 0)
    • Radius: r1=3r_1 = 3

  2. Circle C2C_2: Equation: (x−3)2+(y−4)2=16\text{Equation: } (x - 3)^2 + (y - 4)^2 = 16

    • Centre: O2=(3,4)O_2 = (3, 4)
    • Radius: r2=4r_2 = 4

  3. Line of Centres: The line passing through O1(0,0)O_1(0, 0) and O2(3,4)O_2(3, 4) has direction cosines: cos⁡θ=35,sin⁡θ=45\cos\theta = \frac{3}{5}, \quad \sin\theta = \frac{4}{5} The parametric coordinates of any point on this line can be written as (scos⁡θ,ssin⁡θ)(s\cos\theta, s\sin\theta), where ss represents the directed distance from the origin O1O_1.

    • For circle C1C_1, its centre is at s=0s = 0, so it spans the interval s∈[−3,3]s \in [-3, 3].
    • For circle C2C_2, its centre is at s=5s = 5, so it spans the interval s∈[5−4,5+4]=[1,9]s \in [5 - 4, 5 + 4] = [1, 9].

  4. Circle C3C_3: Since C1C_1 and C2C_2 both lie inside C3C_3, and C3C_3 touches C1C_1 at MM and C2C_2 at NN:

    • The point of contact MM corresponds to the minimum ss-coordinate: sM=−3s_M = -3.
    • The point of contact NN corresponds to the maximum ss-coordinate: sN=9s_N = 9.

    Thus, the segment MNMN is a diameter of C3C_3: Diameter of C3=sN−sM=9−(−3)=12  ⟹  Radius r=6\text{Diameter of } C_3 = s_N - s_M = 9 - (-3) = 12 \implies \text{Radius } r = 6 The centre O3(h,k)O_3(h, k) is the midpoint of MNMN: sO3=−3+92=3s_{O_3} = \frac{-3 + 9}{2} = 3 Therefore, the coordinates of the centre are: h=3cos⁡θ=3(35)=95h = 3\cos\theta = 3 \left(\frac{3}{5}\right) = \frac{9}{5} k=3sin⁡θ=3(45)=125k = 3\sin\theta = 3 \left(\frac{4}{5}\right) = \frac{12}{5}

Step 2: Evaluate the items in List-I

  • Item (I): 2h+k=2(95)+125=18+125=305=62h + k = 2\left(\frac{9}{5}\right) + \frac{12}{5} = \frac{18 + 12}{5} = \frac{30}{5} = 6 Thus, (I)→(P)\text{(I)} \to \text{(P)}.

  • Item (II): The line passing through XX and YY is the radical axis (common chord) of circles C1C_1 and C2C_2: C1−C2=0  ⟹  (x2+y2−9)−(x2+y2−6x−8y+9)=0C_1 - C_2 = 0 \implies (x^2 + y^2 - 9) - (x^2 + y^2 - 6x - 8y + 9) = 0 6x+8y−18=0  ⟹  3x+4y−9=06x + 8y - 18 = 0 \implies 3x + 4y - 9 = 0

    1. Length of XYXY: Distance from O1(0,0)O_1(0, 0) to the line 3x+4y−9=03x + 4y - 9 = 0: d1=∣3(0)+4(0)−9∣32+42=95d_1 = \frac{|3(0) + 4(0) - 9|}{\sqrt{3^2 + 4^2}} = \frac{9}{5} Length of XY=2r12−d12=232−(95)2=29−8125=2×125=245\text{Length of } XY = 2\sqrt{r_1^2 - d_1^2} = 2\sqrt{3^2 - \left(\frac{9}{5}\right)^2} = 2\sqrt{9 - \frac{81}{25}} = 2 \times \frac{12}{5} = \frac{24}{5}

    2. Length of ZWZW: Distance from O3(95,125)O_3\left(\frac{9}{5}, \frac{12}{5}\right) to the line 3x+4y−9=03x + 4y - 9 = 0: d3=∣3(95)+4(125)−9∣32+42=∣27+485−9∣5=∣15−9∣5=65d_3 = \frac{\left|3\left(\frac{9}{5}\right) + 4\left(\frac{12}{5}\right) - 9\right|}{\sqrt{3^2 + 4^2}} = \frac{\left|\frac{27 + 48}{5} - 9\right|}{5} = \frac{|15 - 9|}{5} = \frac{6}{5} Length of ZW=2r2−d32=262−(65)2=2×61−125=2465\text{Length of } ZW = 2\sqrt{r^2 - d_3^2} = 2\sqrt{6^2 - \left(\frac{6}{5}\right)^2} = 2 \times 6\sqrt{1 - \frac{1}{25}} = \frac{24\sqrt{6}}{5}

    3. Ratio: Length of ZWLength of XY=2465245=6\frac{\text{Length of } ZW}{\text{Length of } XY} = \frac{\frac{24\sqrt{6}}{5}}{\frac{24}{5}} = \sqrt{6} Thus, (II)→(Q)\text{(II)} \to \text{(Q)}.

Conclusion

Matching with the given options:

  • (A) (I), (S) is incorrect.
  • (B) (I), (U) is incorrect.
  • (C) (II), (Q) is correct.
  • (D) (II), (T) is incorrect.

The only correct combination is (C).

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