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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2019 · 12 January · Morning Shift · Mathematics · Circle

If a variable line 3x + 4y - λ = 0 is such that the two circles x ^...

If a variable line 3x+4y−λ=03x + 4y - \lambda = 0 is such that the two circles x2+y2−2x−2y+1=0{x}^{2} + {y}^{2} - 2x - 2y + 1 = 0 and x2+y2−18x−2y+78=0{x}^{2} + {y}^{2} - 18x - 2y + 78 = 0 are on its opposite sides, then the set of all values of λ\lambda is the interval :

Options

  1. A[13,23]\left[13, 23\right]
  2. B(23,31)\left(23, 31\right)
  3. C[12,21]\left[12, 21\right]
  4. D(2,17)\left(2, 17\right)

Correct answer

C. [12,21]\left[12, 21\right]

Step-by-step solution

Given circles x2+y2−2x−2y+1=0⇒(x−1)2+(y−1)2=12{x}^{2} + {y}^{2} - 2x - 2y + 1 = 0 ⇒ {\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = {1}^{2} and

x2+y2−18x−2y+78=0⇒(x−9)2+(y−1)2=22{x}^{2} + {y}^{2} - 18x - 2y + 78 = 0 ⇒ {\left(x - 9\right)}^{2} + {\left(y - 1\right)}^{2} = {2}^{2}

Thus, centers of two circles are \left(1, 1\right) \text{ and } \left(9, 1\right)andradiiareand radii are1 \text{ and } 2respectively.</p><p>Since,centerofcirclesareontheoppositesideoflinerespectively.</p><p>Since, center of circles are on the opposite side of line3x + 4y - \lambda = 0.</p><p>.</p><p>⇒ \left(3 + 4 - \lambda\right)\left(27 + 4 - \lambda\right) < 0</p><p></p><p>⇒ \left(7 - \lambda\right)\left(31 - \lambda\right) < 0</p><p></p><p>⇒ \lambda \in \left(7, 31\right) \ldots \ldots . \left(i\right)<br/><br/>Since,circlesshouldnotintersectgivenline.</p><p>Hence,∣7−λ32+42∣≥1 and  ∣31−λ32+42∣≥2<br /><br />Since, circles should not intersect given line.</p><p>Hence, \left|\dfrac{7 - \lambda}{\sqrt{{3}^{2} + {4}^{2}}}\right| \ge 1 \text{\text{ and } } \left|\dfrac{31 - \lambda}{\sqrt{{3}^{2} + {4}^{2}}}\right| \ge 2

⇒∣7−λ∣≥5 and  ∣31−λ∣≥10⇒ \left|7 - \lambda\right| \ge 5 \text{\text{ and } } \left|31 - \lambda\right| \ge 10

⇒λ≤2⇒ \lambda \le 2 or λ≥12…….(ii)\lambda \ge 12 \ldots \ldots . \left(ii\right)

and λ≤21\lambda \le 21 or λ≥41…….(iii)</p><p>Thus,\lambda \ge 41 \ldots \ldots .\left(iii\right)</p><p>Thus,\lambda \in \left(i\right) \cap \left(ii\right) \cap \left(iii\right)

⇒λ∈[12,21]⇒ \lambda \in \left[12,21\right]

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