Circle of radius of Intersection of Plane and Sphere The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius?
I tried putting value of y from plane in sphere but then I get a $zx$ term. How to proceed?

First a normal vector to the plane is $(1,2,-1)$. The sphere equation could be rewritten as $\left(x-\dfrac12\right)^2+y^2+\left(z+\dfrac12\right)^2=\dfrac52$, hence the center is $\left(\dfrac12,0,\dfrac{-1}2\right)$. The center of the intersection circle is on the line $\left(\dfrac12+t,2t,\dfrac{-1}2-t\right)$, satisfying $x+2y-z=4$, that is 
$\dfrac12+t+4t+\dfrac12+t=4$ from which you get $6t=3$, $t=\dfrac12$. Thus the 
center of the intersection circle is $(1,1,-1)$ and from here with a little(hmm?) more effort you could find the equation of the circle. 
Perhaps easier (given you are only looking for the radius, but not for the equation of the intersection circle), do not try to find the actual equation of the intersection circle (which might be tedious) but replace the sphere with one through the origin, and replace the plane with a horizontal plane (i.e. perpendicular to the $z$ axis). As long as the sphere has the same radius, and the center of the intersecting circle is the same distance from the center of the sphere, then the radius of the intersection circle would be the same.  
Note that the distance of the two centers in question is $\sqrt{\left(\dfrac12\right)^2+1^2+\left(\dfrac12\right)^2}=\sqrt{\dfrac32}$. Consider the new sphere $x^2+y^2+z^2=\dfrac52$ and new plane $z=\sqrt{\dfrac32}$. Substituting $z$ from the equation of the (new) plane into the (new) sphere gives you the equation $x^2+y^2=1$, which radius is, of course, $1$. 
