JEE Main 2025 — 2 April, Morning Shift — Question 46

$\because \quad x+y=3$ and $x-y=3$ are tangents

$\therefore \quad$ Both circle centre will lie on $x$-axis

$\therefore \quad(x-a)^{2}+y^{2}=r^{2}$

Hence centre is $C(\alpha, 0)$

$$\begin{gathered} r=\sqrt{(\alpha+9)^{2}+16} …(1) \end{gathered}$$

Also $\left|\frac{\alpha-3}{\sqrt{2}}\right|=r …(2)$

$\sqrt{(\alpha+9)^{2}+16}=\left|\frac{\alpha-3}{\sqrt{2}}\right|$

$\Rightarrow \quad \alpha=-5$ or -37

$r=\left|\frac{-5-3}{\sqrt{2}}\right|$ or $\left|\frac{-37-3}{\sqrt{2}}\right|$

$=4 \sqrt{2}$ or $20 \sqrt{2}$

$\left|r_{1}^{2}-r_{2}^{2}\right|=|32-800|=768$