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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Mathematics
Chapter
Circle

JEE Main 2014 · 11 April · Online · Mathematics · Circle

The set of all real values of λ for which exactly two common...

The set of all real values of λ\lambda for which exactly two common tangents can be drawn to the circles x2+y2−4x−4y+6=0x^2+y^2-4 x-4 y+6=0 and x2+y2−10x−10y+λ=0\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+\lambda=0 is the interval:

Options

  1. A(12,32)(12,32)
  2. B(18,42)(18,42)
  3. C(12,24)(12,24)
  4. D(18,48)(18,48)

Correct answer

B. (18,42)(18,42)

Step-by-step solution

The equations of the circles are x^2+y^2-10 x-10 y+\lambda=0 and x2+y2−4x−4y+6=0x^2+y^2-4 x-4 y+6=0 C1= centre of (1)=(5,5)C2= centre of (2)=(2,2)d= distance between centres =C1C2=9+9=18r1=50−λ,r2=2 For exactly two common tangents we  have r1−r2<C1C2<r1+r2⇒50−λ−2<32<50−λ+2⇒50−λ−2<32 or 32<50−λ+2⇒50−λ<42 or 22<50−λ⇒50−λ<32 or 8<50−λ⇒λ>18 or λ<42 Required interval is (18,42)\begin{aligned} &C_1=\text{ centre of }(1)=(5,5) \\ &\mathrm{C}_2=\text{ centre of }(2)=(2,2) \\ &d=\text{ distance between centres } \\ &=\mathrm{C}_1 \mathrm{C}_2=\sqrt{9+9}=\sqrt{18} \\ &r_1=\sqrt{50-\lambda}, r_2=\sqrt{2} \\ &\text{ For exactly two common tangents we } \\ &\text{ have } \\ &r_1-r_2 < C_1 C_2 < r_1+r_2 \\ &\Rightarrow \sqrt{50-\lambda}-\sqrt{2} < 3 \sqrt{2} < \sqrt{50-\lambda}+\sqrt{2} \\ &\Rightarrow \sqrt{50-\lambda}-\sqrt{2} < 3 \sqrt{2} \text{ or } 3 \sqrt{2} < \sqrt{50-\lambda}+\sqrt{2} \\ &\Rightarrow \sqrt{50-\lambda} < 4 \sqrt{2}\text{ or } 2 \sqrt{2} < \sqrt{50-\lambda} \\ &\Rightarrow 50-\lambda < 32\text{ or } 8 < 50-\lambda \\ &\Rightarrow \lambda>18\text{ or } \lambda < 42 \\ &\text{ Required interval is }(18,42) \\ & \end{aligned}
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