Given a circle $5=0$ whose centre lies on the line $\mathrm{x}+\mathrm{y}-5=0$
Let the centre of circle -be,$(h, k)$



So, $(h, 5)$ and $(2, k)$ also
Satisfy the equation of circle.
Equation of circle is $(x-4)^2+(y-k)^2=r^2$.
Satisfy the points (h, 5) and $(2, \mathrm{k})$
$$
\begin{aligned}
& (\mathrm{h}-\mathrm{h})^2+(5-\mathrm{k})^2=\mathrm{r}^2 \\
& 5-\mathrm{k}=\mathrm{r} \ldots \ldots\left(\mathrm{a}^{\prime}\right) \\
& \left.(2=\mathrm{h})^2+\mathrm{k}-\mathrm{k}\right)^2=\mathrm{r}^2 \\
& 2=\mathrm{h}=\mathrm{r} \ldots \ldots . .\left(\mathrm{b}^{\prime}\right) \\
& \text { Here, }\left(\mathrm{a}^{\prime}\right)=\left(\mathrm{b}^{\prime}\right) \\
& 5-\mathrm{k}=2-\mathrm{h}
\end{aligned}
$$

Add (i) \& (ii)
$$
\begin{aligned}
& \mathrm{h}+\mathrm{k}=5 \\
& \frac{\mathrm{h}-\mathrm{k}=-3}{2 \mathrm{~h}=2} \\
& \mathrm{~h}=1
\end{aligned}
$$
from (i)
$$
\mathrm{h}+\mathrm{k}=5
$$
$$
\mathrm{k}=4
$$
Then, $(\mathrm{h}, \mathrm{k}) \rightarrow(1,4)$
from $\left(a^{\prime}\right)$
$$
\Rightarrow \mathrm{r}=5-4=1
$$
Required area of circle $=\pi \mathrm{r}^2=\pi(1)^2=\pi$
So, option (a) is correct