Does the sphere $x^2 + (y-2)^2 + z^2 = 1$ intersect the plane $z-x = 3$? If so, find a point in their intersection. If not explain why. 
Does the sphere $x^2 + (y-2)^2 + z^2 = 1$ intersect the plane $z-x = 3$? If so, find a point in their intersection. If not explain why.

I feel like I can give a convincing answer with a rough sketch, and words, but I can not figure out how to solve the problem algebraically. If anyone could help I would be greatly appreciative!

Let consider the translation $X=x$, $Y=y-2$, $Z=z$ then

*

*$x^2+(y-2)^2+z^2=1 \iff X^2+Y^2+Z^2=1$

*$x-z=3 \iff X-Z=3$
then consider the circle and the line in the plane $Y=0$

*

*$X^2+Z^2=1$

*$X-Z=3$
