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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Circle

Let the set of all values of r , for which the circles (x+1)^ 2...

Let the set of all values of rr, for which the circles (x+1)2+(y+4)2=r2(x+1)^{2}+(y+4)^{2}=r^{2} and x2+y2−4x−2y−4=0x^{2}+y^{2}-4 x-2 y-4=0 intersect at two distinct points be the interval (α,β)(\alpha, \beta). Then αβ\alpha \beta is equal to

Options

  1. A21
  2. B24
  3. C20
  4. D25

Correct answer

D. 25

Step-by-step solution

Circle 1: center C1=(−1,−4)C_1 = (-1, -4), radius rr.
Circle 2: (x−2)2+(y−1)2=9(x-2)^2 + (y-1)^2 = 9, center C2=(2,1)C_2 = (2, 1), radius 33.
Distance d=9+25=34d = \sqrt{9+25} = \sqrt{34}.
For two intersection points: ∣r−3∣<34<r+3|r-3| < \sqrt{34} < r+3.
r+3>34⇒r>34−3r + 3 > \sqrt{34} \Rightarrow r > \sqrt{34}-3 and ∣r−3∣<34⇒r<34+3|r-3| < \sqrt{34} \Rightarrow r < \sqrt{34}+3.
r∈(34−3, 34+3)r \in (\sqrt{34}-3,\ \sqrt{34}+3), so α=34−3\alpha = \sqrt{34}-3, β=34+3\beta = \sqrt{34}+3.
αβ=(34)2−9=34−9=25\alpha\beta = (\sqrt{34})^2 - 9 = 34 - 9 = 25.
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