Mathematics · 2024
JEE Main · 27 January 2024, Shift 2 · Q28
Consider a circle (x-α)^2+(y-β)^2=50, where α, β>0. If the circle touches the line y+x=0 at the point P, whose distance from the origin is 4 √ 2,…
Consider a circle $\displaystyle (x-\alpha)^2+(y-\beta)^2=50$, where $\displaystyle \alpha, \beta>0$. If the circle touches the line $\displaystyle y+x=0$ at the point P, whose distance from the origin is $\displaystyle 4 \sqrt{2}$, then $\displaystyle (\alpha+\beta)^2$ is equal to $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
100
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.