The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius:

Sphere: $x^2+y^2+z^2-x+z-2=0$.Centre $C=(1/2,0,-1/2)$.Radius $R=\sqrt{(1/2{)}^2+0^2+(-1/2{)}^2-(-2)}=\sqrt{1/4+1/4+2}=\sqrt{2.5}=\sqrt{5/2}$.Perpendicular distance $p$ from $C$ to the plane $x+2y-z-4=0$:$p=\frac{|(1/2)+2(0)-(-1/2)-4|}{\sqrt{1^2+2^2+(-1{)}^2}}$$p=\frac{|1/2+1/2-4|}{\sqrt{6}}=\frac{|-3|}{\sqrt{6}}=\frac{3}{\sqrt{6}}=\sqrt{\frac{9}{6}}=\sqrt{\frac{3}{2}}$.Radius of the circle $r=\sqrt{R^2-p^2}$:$r=\sqrt{5/2-3/2}=\sqrt{2/2}=1$.