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कृपया अपनी पसंदीदा भाषा चुनें
एक वृत्त $(x-\alpha)^{2}+(y-\beta)^{2}=50$ को विचार करें, जहाँ $\alpha, \beta>0$। यदि वृत्त रेखा $y+x=0$ को बिंदु $P$ पर स्पर्श करता है, जिसकी मूल बिंदु से दूरी $4 \sqrt{2}$ है, तो $(\alpha+\beta)^{2}$ के बराबर है
$S:(x-\alpha)^{2}+(y-\beta)^{2}=50$
$C P=r$
$\left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2}$
$\Rightarrow(\alpha+\beta)^{2}=100$
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