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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Circle

Let C_1 be the circle in the third quadrant of radius 3 , that...

Let C1C_1 be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let C2\mathrm{C}_2 be the circle with centre (1,3)(1,3) that touches C1\mathrm{C}_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 :

Options

  1. A9
  2. B13
  3. C22
  4. D31

Correct answer

C. 22

Step-by-step solution

C_1:(x+3)^2+(y+3)^2=3^2Let C1\mathrm{C}_1 and C2\mathrm{C}_2 has centres A(−31−3)A\left(-3_1-3\right) and B(1,3)B(1,3)AB=16+36=213r1=3 and r2=213−3P(α,β),α=r1(1)+r2(−3)r1+r2,β=r1(3)+r2(−3)r1+r2α=3−3(213−3)213,β=18−613213(β−α)2=(6213)2(β−α)2=(6213)2,m+n=22\begin{aligned}& \mathrm{AB}=\sqrt{16+36}=2 \sqrt{13} \\ & \mathrm{r}_1=3 \text{ and } \mathrm{r}_2=2 \sqrt{13}-3 \\ & \mathrm{P}(\alpha, \beta), \alpha=\frac{\mathrm{r}_1(1)+\mathrm{r}_2(-3)}{\mathrm{r}_1+\mathrm{r}_2}, \beta=\frac{\mathrm{r}_1(3)+\mathrm{r}_2(-3)}{\mathrm{r}_1+\mathrm{r}_2} \\ & \alpha=\frac{3-3(2 \sqrt{13}-3)}{2 \sqrt{13}}, \beta=\frac{18-6 \sqrt{13}}{2 \sqrt{13}} \\ & (\beta-\alpha)^2=\left(\frac{6}{2 \sqrt{13}}\right)^2 \\ & (\beta-\alpha)^2=\left(\frac{6}{2 \sqrt{13}}\right)^2, \mathrm{ m}+\mathrm{n}=22\end{aligned}$$
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