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!
 A: If so, $$x^2+(y-2)^2+(x+3)^2=1$$ or
$$2(x^2+3x+4)+(y-2)^2=0,$$ which is impossible.
A: 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 $Y=0$

*

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

*$X-Z=3$
A: Here's a geometric/algebraic answer. The well known formula for the distance from a point $P=(x_0,y_0,z_0)$ to a plane $Ax+By+Cz+D=0$ is $$d=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$$.
So the distance from the center of the sphere $(0,2,0)$ to the plane $x-z+3=0$ is
$$d=\frac{|1\cdot 0+0\cdot 2+(-1)\cdot 0+3|}{\sqrt{1^2+0^2+(-1)^2}} = \frac{3}{\sqrt{2}}$$
So the distance from the center of the sphere to the plane $\frac{3}{\sqrt{2}}$ is greater than the radius of the sphere $1$. Then the sphere doesn't intersect the plane.
A: No. To see this notice the plane can be written at $z=x+3$. Substituting into the equation of the sphere yields:
$$ x^2 + (y-2)^2 + (x+3)^2 = 1 $$
When we expand the last term of the left hand side this gives us:
$$ 2x^2 + 6x+ (y-2)^2 = -8$$
Completing the square then gives us:
$$2\left(x+\frac{3}{2}\right)^2 + (y-2)^2 = -8+\frac{9}{2}= \frac{-7}{2}$$
Notice that the two terms on the righthand side are always greater than or equal to zero, but the lefthand side is negative. Thus, for the reals, there is no intersection.
