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

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

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

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

$$\begin{aligned} &r=\sqrt{(\alpha+9)^2+16}\quad\text{.... (1)}\\ &\text { Also }\left|\frac{\alpha-3}{\sqrt{2}}\right|=r \quad\text{.... (2)}\\ &\begin{aligned} & \sqrt{(\alpha+9)^2+16}=\left|\frac{\alpha-3}{\sqrt{2}}\right| \\ & \Rightarrow \quad \alpha=-5 \text { or }-37 \\ & \mathrm{r}=\left|\frac{-5-3}{\sqrt{2}}\right| \text { or }\left|\frac{-37-3}{\sqrt{2}}\right| \\ & =4 \sqrt{2} \text { or } 20 \sqrt{2} \\ & \left|\mathrm{r}_1^2-\mathrm{r}_2^2\right|=|32-800|=768 \end{aligned} \end{aligned}$$

19:T56a,

JEE Main 2025 (Online) 2nd April Morning Shift Mathematics - Circle Question 19 Hindi Explanation

$\because x+y=3$ और $x-y=3$ स्पर्श रेखाएँ हैं

$\therefore \quad$ दोनों वृत्तों के केंद्र $x$-अक्ष पर स्थित होंगे

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

अतः केंद्र $C(\alpha, 0)$ है

$$\begin{aligned} &r=\sqrt{(\alpha+9)^2+16}\quad\text{.... (1)}\\ &\text { इसके अतिरिक्त }\left|\frac{\alpha-3}{\sqrt{2}}\right|=r \quad\text{.... (2)}\\ &\begin{aligned} & \sqrt{(\alpha+9)^2+16}=\left|\frac{\alpha-3}{\sqrt{2}}\right| \\ & \Rightarrow \quad \alpha=-5 \text { या

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