Question

The least area of a circle circumscribing any right triangle of area $S$ is

• A. $\pi S$
• B. $2\pi S$
• C. $\sqrt{2}\pi S$
• D. $4\pi S$

Answer 1

Answer: (A)

$S=\frac{xy}{2}=\text{ constant }$
$\text{ Area of the circles }(A)=\pi r^2=\frac{\pi \left(x^2+y^2\right)}{4};\left(x^2+y^2=4r^2\right)$
$A(x)=\frac{\pi }{4}\left[x^2+{\left(\frac{2S}{x}\right)}^2\right]$
$A^{\prime }(x)=2x-\frac{8s^2}{x^3}=0\Rightarrow x^4=4S^2\Rightarrow x^2=2S$
$S^2=\frac{x^2{y}^2}{4}=\frac{2Sy}{4}\Rightarrow y^2=2S$
$\therefore \text{ least area of circle }=\pi r^2=\frac{\pi }{4}\left(x^2+y^2\right)=\pi S$