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?

As one answer already pointed out,  first complete the squares to simplify the equation of the sphere:
$$(x-0.5)^2+y^2+(z+0.5)^2=2.5$$
So the centre of the sphere is at $(0.5,0,-0.5)$ and its radius is $R=\sqrt{\dfrac 52}$
Now say the distance of the plane from the centre of sphere is $d$. Simply applying the necessary formula we get $d=\sqrt{\dfrac 32}$
Now just draw a diagram of the situation and you'll find that the radius of the circle of intersection $r$ is simply $$r=\sqrt{R^2-d^2}=1$$
