If in a triangle $ABC,$ right angled at $B, s-a=3, s-c=2,$ then $a^2+c^2=$

We have,

$s-a=3\Rightarrow b+c-a=6$                          (i)

and, $s-c=2\Rightarrow a+b-c=4$                  (ii)

Adding these two equations, we get $b=5$

Since $B$ is a right angle

$\therefore b^2=a^2+c^2\Rightarrow a^2+c^2=25$

Multiplying (i) and (ii), we get

$\begin{array}{l}\left[\right(b-c)+a]\left[\right(b+c)-a]=24\\ \Rightarrow b^2-c^2+2ac-a^2=24\\ \Rightarrow a^2+2ac-a^2=24\left[\because b^2=a^2+c^2\right] \left[\because b^2=a^2+c^2\right]\\ \Rightarrow ac=12 (\text{iv)}\end{array}$

From (iii) and (iv), we have

$a+c=7$ and $a-c=1\Rightarrow a=4,c=3\Rightarrow a^2+c^2=25$