The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing though the points $A$ and $B$ and having least radius is

$\begin{array}{l}x-2=0,x^2+y^2-8x-2y+8=0\Rightarrow 4+y^2-16-2y+8=0\Rightarrow y^2-2y-4=0\\ \Rightarrow y=\frac{2\pm \sqrt{4+16}}{2}=1\pm \sqrt{5}.\therefore A=(2,1+\sqrt{5}),B=(2,1-\sqrt{5})\end{array}$

Circle through $A, B$ and having least radius  is  the circle having $AB$ as diameter.

$\therefore$ It's equation is $(x-2)(x-2)+(y-1-\sqrt{5})(y-1+\sqrt{5})=0$

$\Rightarrow x^2-4x+4+(y-1{)}^2-5=0\Rightarrow x^2+y^2-4x-2y=0.$