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

Given that the least area of a circle is circumscribing any right triangle of area 

It is required to find the area of $S$

Assume that the sides of the triangle are $a,b,2r$

$Area=S=\frac{1}{2}\times$ Base $\times$ Height

$S=\frac{ab}{2}$

Hence, $b=\frac{2S}{a}$ and the radius is,

$(2r{)}^2=a^2+b^2$

$$\begin{gathered} 4r^2=a^2+b^2 \\ {r}^2=\frac{a^2+b^2}{4} \end{gathered}$$

Evaluate the area of the circle.

$$\begin{gathered} A=\pi r^2 \\ A=\frac{\pi }{4}\left(a^2+b^2\right) \\ A=\frac{\pi }{4}\left(a^2+{\left(\frac{2S}{a}\right)}^2\right) \\ A=\frac{\pi }{4}\left(a^2+\frac{4S^2}{a^2}\right) \end{gathered}$$

Differentiate with respect to $a$

$$\begin{gathered} \frac{dA}{da}=\frac{\pi }{4}\left(2a-\frac{8S^2}{a^3}\right) \\ \frac{dA}{da}=\frac{\pi }{4}\left(\frac{2a^4-8S^2}{a^3}\right) \end{gathered}$$

The minima are obtained where $\frac{dA}{da}=0$

$$\begin{gathered} \frac{\pi }{4}\left(\frac{2a^4-8S^2}{a^3}\right)=0 \\ 2a^4-8S^2=0 \\ 2\left(a^2-2S\right)\left(a^2+2S\right)=0 \\ {a}^2=2S \\ \frac{a^2}{2}=S \end{gathered}$$

Evaluate the value of $b$.

$b=\frac{2S}{a}$

$b=\frac{a^2}{a}$

$b=a$

Then, the area of the circle is given by

$A=\frac{\pi }{4}\left(a^2+b^2\right)$

$A=\frac{\pi }{4}(2S+2S)$

$A=\frac{4\pi }{4}S$

$A=\pi S$

Hence, If the least area of a circle circumscribing any right triangle of area then $S$ is equal to $\pi s$.

Therefore, the correct answer is option $\left(1\right)$.