JEE MAIN - Mathematics (2025 - 7th April Morning Shift - No. 2)
Let $C_1$ be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let $C_2$ be the circle with centre $(1,3)$ that touches $\mathrm{C}_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}$ , $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to
22
13
9
31
Explanation
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$$ \begin{aligned} & \sqrt{16+36}=r+3 \\ & \Rightarrow r=\sqrt{52}-3 \\ & \Rightarrow \frac{3-3 r}{-r+3}=\alpha, \beta=\frac{9-3 r}{3+r} \\ & \Rightarrow \frac{(9-3 r-3+3 r)^2}{(r+3)^2} \\ & \Rightarrow \frac{36}{12}=\frac{9}{13}=\frac{m}{n} \Rightarrow m+n=22 \end{aligned}$$
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