The equation of the image of the circle ${(x-3)}^2+{(y-2)}^2=1$ by the mirror $x+y=19$ is  

The image of the circle has same radius but centre different. If centre is $(\alpha ,\beta )$ then

$\frac{\alpha -3}{1}=\frac{\beta -2}{1}=\frac{-2(3+2-19)}{1^2+1^2}\Rightarrow \alpha -3=\beta -2=14$

$\therefore \alpha =17,\beta =16$

$\therefore$ Required circle is ${(x-17)}^2+{(y-16)}^2=1$