JEE Challenger
More from Conic Sections
MathHardAdv 2019 • Paper 2

Incorrect Combination for Circle Chord and Parabola Tangency

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 INCORRECT combination?

Options

A

(I), (P)

B

(IV), (U)

C

(III), (R)

D

(IV), (S)

Correct

Topics & Concepts

Conic SectionsCirclesParabolaTangents and Normals

Step-by-Step Solution

To find the incorrect combination, we determine the values corresponding to each item in List-I:

1. Circles C1,C2C_1, C_2 and Determination of C3C_3:

The given circles are:

  • C1:x2+y2=9C_1: x^2 + y^2 = 9, with centre O1=(0,0)O_1 = (0, 0) and radius r1=3r_1 = 3.
  • C2:(x−3)2+(y−4)2=16C_2: (x - 3)^2 + (y - 4)^2 = 16, with centre O2=(3,4)O_2 = (3, 4) and radius r2=4r_2 = 4.

The distance between the centres is: O1O2=32+42=5O_1O_2 = \sqrt{3^2 + 4^2} = 5

The line passing through the centres O1O_1 and O2O_2 has the equation: y=43x  ⟹  4x−3y=0y = \frac{4}{3}x \implies 4x - 3y = 0

A point on this line can be parameterized as: P(s)=(35s,45s)P(s) = \left(\frac{3}{5}s, \frac{4}{5}s\right) where ss is the directed distance from O1(0,0)O_1(0,0) towards O2(3,4)O_2(3,4).

  • For C1C_1, the endpoints of its diameter along this line correspond to s=−3s = -3 and s=3s = 3.
  • For C2C_2, the centre is at s=5s = 5, so the endpoints of its diameter along this line correspond to s=5−4=1s = 5 - 4 = 1 and s=5+4=9s = 5 + 4 = 9.

Since C3C_3 touches C1C_1 internally at MM and C2C_2 internally at NN, and both C1C_1 and C2C_2 lie inside C3C_3, the contact points MM and NN must be the extreme endpoints along the line of centres:

  • Point MM corresponds to s=−3s = -3, so: M=(−95,−125)M = \left(-\frac{9}{5}, -\frac{12}{5}\right)
  • Point NN corresponds to s=9s = 9, so: N=(275,365)N = \left(\frac{27}{5}, \frac{36}{5}\right)

The diameter of C3C_3 is the segment MNMN: Diameter=9−(−3)=12  ⟹  Radius r=6\text{Diameter} = 9 - (-3) = 12 \implies \text{Radius } r = 6

The centre of C3C_3 is the midpoint of MM and NN, which corresponds to s=−3+92=3s = \frac{-3 + 9}{2} = 3: O3=(h,k)=(95,125)O_3 = (h, k) = \left(\frac{9}{5}, \frac{12}{5}\right)

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

2. Length of Chords XYXY and ZWZW:

The line passing through the intersection points XX and YY of C1C_1 and C2C_2 is their radical axis: C1−C2=0  ⟹  (x2+y2−9)−((x−3)2+(y−4)2−16)=0C_1 - C_2 = 0 \implies (x^2 + y^2 - 9) - ((x-3)^2 + (y-4)^2 - 16) = 0 6x+8y−18=0  ⟹  3x+4y−9=06x + 8y - 18 = 0 \implies 3x + 4y - 9 = 0

The distance of the line 3x+4y−9=03x + 4y - 9 = 0 from O1(0,0)O_1(0,0) is: d1=∣−9∣32+42=95d_1 = \frac{|-9|}{\sqrt{3^2 + 4^2}} = \frac{9}{5}

The length of chord XYXY is: Length of XY=2r12−d12=232−(95)2=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 \times \frac{12}{5} = \frac{24}{5}

The distance of the line 3x+4y−9=03x + 4y - 9 = 0 from the centre O3(95,125)O_3\left(\frac{9}{5}, \frac{12}{5}\right) is: d3=∣3(95)+4(125)−9∣32+42=∣755−9∣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{75}{5} - 9\right|}{5} = \frac{6}{5}

The length of chord ZWZW is: Length of ZW=2r2−d32=262−(65)2=2×1265=2465\text{Length of } ZW = 2\sqrt{r^2 - d_3^2} = 2\sqrt{6^2 - \left(\frac{6}{5}\right)^2} = 2 \times \frac{12\sqrt{6}}{5} = \frac{24\sqrt{6}}{5}

Therefore: 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} Hence, (II) corresponds to (Q).

3. Ratio of Areas of △MZN\triangle MZN and △ZMW\triangle ZMW:

The line MNMN has direction (3,4)(3, 4), which is perpendicular to the line of the chord ZWZW (3x+4y=93x + 4y = 9). Thus, the diameter MNMN is the perpendicular bisector of chord ZWZW.

Let HH be the point of intersection of MNMN and ZWZW:

  • HH lies on MNMN, so its parameter ss satisfies 3(3s5)+4(4s5)=9  ⟹  s=953\left(\frac{3s}{5}\right) + 4\left(\frac{4s}{5}\right) = 9 \implies s = \frac{9}{5}.
  • The length of the altitude from MM (where s=−3s = -3) to ZWZW is: MH=∣95−(−3)∣=245MH = \left|\frac{9}{5} - (-3)\right| = \frac{24}{5}

The areas of the triangles are:

  • For △MZN\triangle MZN, the base is MN=12MN = 12, and the altitude from ZZ to MNMN is 12ZW\frac{1}{2}ZW: Area(△MZN)=12×MN×(12ZW)=14×12×ZW=3 ZW\text{Area}(\triangle MZN) = \frac{1}{2} \times MN \times \left(\frac{1}{2}ZW\right) = \frac{1}{4} \times 12 \times ZW = 3\,ZW
  • For △ZMW\triangle ZMW, the base is ZWZW, and the altitude from MM is MH=245MH = \frac{24}{5}: Area(△ZMW)=12×ZW×MH=12×ZW×245=125 ZW\text{Area}(\triangle ZMW) = \frac{1}{2} \times ZW \times MH = \frac{1}{2} \times ZW \times \frac{24}{5} = \frac{12}{5}\,ZW

Therefore: Area of triangle MZNArea of triangle ZMW=3 ZW125 ZW=1512=54\frac{\text{Area of triangle } MZN}{\text{Area of triangle } ZMW} = \frac{3\,ZW}{\frac{12}{5}\,ZW} = \frac{15}{12} = \frac{5}{4} Hence, (III) corresponds to (R).

4. Value of α\alpha:

Since C1C_1 and C3C_3 touch internally at MM, their common tangent is the tangent at M(−95,−125)M\left(-\frac{9}{5}, -\frac{12}{5}\right): 3(x+95)+4(y+125)=0  ⟹  3x+4y+15=0  ⟹  y=−34x−1543\left(x + \frac{9}{5}\right) + 4\left(y + \frac{12}{5}\right) = 0 \implies 3x + 4y + 15 = 0 \implies y = -\frac{3}{4}x - \frac{15}{4}

A line y=mx+cy = mx + c is tangent to the parabola x2=4Ayx^2 = 4Ay (where A=2αA = 2\alpha) if: c=−Am2c = -Am^2

Substituting m=−34m = -\frac{3}{4} and c=−154c = -\frac{15}{4}: −154=−(2α)(−34)2  ⟹  154=2α×916  ⟹  α=103-\frac{15}{4} = -(2\alpha)\left(-\frac{3}{4}\right)^2 \implies \frac{15}{4} = 2\alpha \times \frac{9}{16} \implies \alpha = \frac{10}{3} Hence, (IV) corresponds to (U).

Conclusion:

  • (A) (I), (P) is CORRECT.
  • (B) (IV), (U) is CORRECT.
  • (C) (III), (R) is CORRECT.
  • (D) (IV), (S) is INCORRECT because (IV) matches with (U).

Thus, the only incorrect combination is (D).

Previous