Circle Question 5
Question 5 - 2024 (27 Jan Shift 2)
Consider a circle $(x-\alpha)^{2}+(y-\beta)^{2}=50$, where $\alpha, \beta>0$. If the circle touches the line $y+x=0$ at the point $\mathrm{P}$, whose distance from the origin is $4 \sqrt{2}$, then $(\alpha+\beta)^{2}$ is equal to
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Answer (100)
Solution
$S:(x-\alpha)^{2}+(y-\beta)^{2}=50$
$C P=r$
$\left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2}$
$\Rightarrow(\alpha+\beta)^{2}=100$
