Jee Main 2024 27 01 2024 Shift 2 - Question16
Question 16
If the line $x+y=0$ is tangent to the circle $(x-\lambda)^{2}+$ $(y-\beta)^{2}=50$, then $(\lambda+\beta)^{2}=$
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Answer: (100.00)
Solution:
Perpendicular distance from centre $=$ radius
$$ \begin{gathered} \Rightarrow \quad\left|\frac{\lambda+\beta}{\sqrt{2}}\right|=\sqrt{50} \\ \quad|\lambda+\beta|=\sqrt{100} \\ (\lambda+\beta)^{2}=100 \end{gathered} $$
BETA
