The equation of the image of the circle $x^{2}+y^{2}-6x-4y+12=0$ by the line ;mirror $x+y-1=0$ is

The image of circle w.r.t same line means image of centre w.r.t that line without changing radius

$x^{2}+y^{2}-6x-4y+12=0$

Centre $=(3,2)$ Radius$=1$

Image of $(3,2)$ w.r.t $x+y-1=0$

$\dfrac {x-3}{1}=\dfrac {y-2}{1}=-2\dfrac {(3+2-1)}{1^{1}+1^{2}}=-4$

$x=-1$, $y=-2$

Then equation of image of circle is

$(x+1)^{2}+(y+2)^{2}=(1)^{2}$

$\Rightarrow x^{2}+y^{2}+2x+4y+4=0$