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Mathematics · Circles

JEE Main 2026 — 22 January, Morning Shift — Question 12

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\alpha, β\beta ). Then αβ\alpha \beta is equal to

  1. Option A:

    25

    Correct
  2. Option B:

    20

  3. Option C:

    21

  4. Option D:

    24

Answer: A

Step-by-step solution

(x−2)2+(y−1)2=32(x-2)^2 + (y-1)^2 = 3^2 (x+1)2+(y+4)2=r2(x+1)^2 + (y+4)^2 = r^2 ∣r1−r2∣<c1c2<r1+r2|r_1 - r_2| < c_1c_2 < r_1 + r_2 ∣r−3∣<(2+1)2+(1+4)2|r - 3| < \sqrt{(2+1)^2 + (1+4)^2} ∣r−3∣<34|r - 3| < \sqrt{34} −34<r−3<34-\sqrt{34} < r - 3 < \sqrt{34} 3−34<r<3+343 - \sqrt{34} < r < 3 + \sqrt{34} r∈(3−34, 3+34)∩(34−3, ∞)r \in (3 - \sqrt{34},\, 3 + \sqrt{34}) \cap (\sqrt{34} - 3,\, \infty) ∴  r∈(34−3, 34+3)\therefore \; r \in (\sqrt{34} - 3,\, \sqrt{34} + 3) αβ=(34−3)(34+3)\alpha \beta = (\sqrt{34} - 3)(\sqrt{34} + 3) =34−9= 34 - 9 =25= 25

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Mathematics
Chapter
Circles
Topic
System of Two Circles and Common Tangents
Let the set of all values of r , for which the circles (x+1) 2 +(y+4)… | JEE Main 2026 PYQ with Solution · DhiX AI