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?

$x+2y-4=z$ and $x^2+y^2-x=-(z^2+z-2)$
$$x^2+y^2-x=(z+2)(1-z)$$
Substituting, $$x^2+y^2-x=(x+2y-2)(1-(x+2y-4))$$
$$x^2+y^2-x=-(x+2y-2)(x+2y-5)$$
And this way you get an equation in $x(s)$ and $y$(s).
