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Mathematics · Circles

JEE Main 2025 — 7 April, Morning Shift — Question 29

Let C1C_{1} be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let C2C_{2} be the circle

with centre (1,3)(1,3) that touches C1C_{1} externally at the point (α,β)(\alpha, \beta). If (β−α)2=mn,gcd⁡(m,n)=1(\beta-\alpha)^{2}=\frac{m}{n}, \operatorname{gcd}(m, n)=1,

then m+nm+n is equal to

  1. Option A:

    22

    Correct
  2. Option B:

    31

  3. Option C:

    13

  4. Option D:

    9

Answer: A

Step-by-step solution

16+36=r+3\sqrt{16+36}=r+3

⇒r=52−3\Rightarrow r=\sqrt{52}-3

⇒3−3r−r+3=α,β=9−3r3+r\Rightarrow \frac{3-3 r}{-r+3}=\alpha, \beta=\frac{9-3 r}{3+r}

⇒(9−3r−3+3r)2(r+3)2\Rightarrow \frac{(9-3 r-3+3 r)^{2}}{(r+3)^{2}}

⇒3612=913=mn⇒m+n=22\Rightarrow \frac{36}{12}=\frac{9}{13}=\frac{m}{n} \Rightarrow m+n=22

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Morning Shift
Subject
Mathematics
Chapter
Circles
Topic
System of Two Circles and Common Tangents
Let C 1 be the circle in the third quadrant of radius 3 , that… | JEE Main 2025 PYQ with Solution · DhiX AI