Equation of tangent to the circle $x^2+y^2-2 x-2 y+1=0$ perpendicular to $y=x$ is given by

Let the equation of tangent be $y=m x+c \ldots$ (i) Given equation of circle is $x^2+y^2-2 x-2 y+1=0$ Its centre is $(1,1)$ and radius, $r=\sqrt{1^2+1^2-1}=\sqrt{1+1-1}=1$ We know that, if line $y=m x+c$ be a tangent to the circle, then $c= \pm r \sqrt{1+m^2}$ $\therefore c= \pm 1 \sqrt{1+m^2} \ldots$ (ii) Since, the tangent line is perpendicular to $y=x$. $\therefore m \times 1=-1\left(\because m_1 m_2=-1\right)$ $\Rightarrow m=-1$ On putting $m=-1$ in Eq. (ii), we get $c= \pm 1 \sqrt{1+(-1)^2}= \pm \sqrt{1+1}= \pm \sqrt{2}$ On putting the values of $m=-1$ and $c= \pm \sqrt{2}$ in Eq. (i), we get $y=-x \pm \sqrt{2}$