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Q.
The least area of a circle circumscribing any right triangle of area $S$ is
Application of Derivatives
Solution:
$S =\frac{ xy }{2}=\text { constant }$
$\text { Area of the circles }( A )=\pi r ^2=\frac{\pi\left( x ^2+ y ^2\right)}{4} ;\left( x ^2+ y ^2=4 r ^2\right) $
$A ( x )=\frac{\pi}{4}\left[ x ^2+\left(\frac{2 S }{ x }\right)^2\right] $
$A ^{\prime}( x )=2 x -\frac{8 s ^2}{ x ^3}=0 \Rightarrow x ^4=4 S ^2 \Rightarrow x ^2=2 S$
$S ^2=\frac{ x ^2 y ^2}{4}=\frac{2 Sy }{4} \Rightarrow y ^2=2 S $
$\therefore \quad \text { least area of circle }=\pi r ^2=\frac{\pi}{4}\left( x ^2+ y ^2\right)=\pi S $
