Let  $S_1\equiv x^2+y^2-4x-8y+4=0$ and $S_2$  be its image in the line  $y=x.$ 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 circle  $S_1=(2,4)$
centre of circle  $S_2=(4,2)$
Radius of circle $S_1=$  radius of circle  $S_2=4$
$\therefore$  equation of circles $S_2$
${(x-4)}^2+{(y-2)}^2=16$ 
 $\Rightarrow x^2+y^2-8x-4y+4=\mathrm{0....}\left(i\right)$
Equation of circle touching $y=x$  at $(1,1)$  can be taken as 
 ${(x-1)}^2+{(y-1)}^2+\lambda (x-y)=0$
Or,  $x^2+y^2+x(\lambda -2)+y(-\lambda -2)+2=\mathrm{0....}\left(ii\right)$
this is orthogonal to  $S_2$, Hence use orthogonal condition to find  $\lambda$

$\Rightarrow 2\left(\frac{\lambda -2}{2}\right)\cdot (-4)+2\left(\frac{-\lambda -2}{2}\right)\cdot (-2)=4+2$

$\Rightarrow -4\lambda +8+2\lambda +4=6$

Equation of required circle is 

$x^2+y^2+x-5y+2=0$

$\text{ Radius }=\sqrt{\frac{1}{4}+\frac{25}{4}-2}=\sqrt{\frac{26-8}{4}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$