$Let{S}_1=x^2+y^2-4x-8y+4=0$  and its image in the line  $y=x$ is $S_2$  .The radius of the circle touching y = x at (1, 1) and orthogonal to  $S_{2}is\frac{3}{\sqrt{\lambda }}then{\lambda }^2+2=$

: Centre of  $S_1=(2,4)$
Radius of  $S_1=radiusof{S}_2=4$
Centre of circle  $S_2=(4,2)$
 $S_2={(x-4)}^2+{(y-2)}^2=16$
 $=x^2+y^2-8x-4y+4=\mathrm{0.........}\left(1\right)$
Equation of the circle touching y=x at (1, 1) can be taken as  
 ${(x-1)}^2+{(y-1)}^2+\lambda (x-y)=\mathrm{0......}\left(2\right)$
$\begin{array}{l}2\left(\frac{\lambda -2}{2}\right)(-4)+2\left(\frac{-\lambda -2}{2}\right)+2=4+2(\therefore 1and2orthogonal)\\ \Rightarrow \lambda =2\\ {\lambda }^2+2=6\end{array}$