JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 28)
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 $$P$$, whose distance from the origin is $$4 \sqrt{2}$$, then $$(\alpha+\beta)^2$$ is equal to __________.
Answer
100
Explanation
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$$\begin{aligned} & S:(x-\alpha)^2+(y-\beta)^2=50 \\ & C P=r \\ & \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} \\ & \Rightarrow(\alpha+\beta)^2=100 \end{aligned}$$
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