Hi I am currently working on a problem that says the following:
Let $A$ be the point $(0, 1, 0)$ and $B$ be the point $(0, 0, 2)$. Let $S$ be the sphere $$(x-2)^2+y^2+z^2=4$$ Consider the family of all planes that pass through both the points $A$ and $B$. This forms a one-parameter family of planes in the sense that one additional piece of information will specify the plane. One of these planes is of course the $y$-$z$ coordinate plane which has equation $x = 0$. The remaining planes in the family do not pass through the origin and they can be described by the general equation $$ax+2y+z=2$$ Here the single parameter is $a$. Do check that for every $a$, this plane passes through $A$ and $B$. Your job is to find the value of $a$ for which the plane is tangent to the sphere $S$. Do not use calculus.
My attempt: I realized that by looking at the sphere and plane from a certain perspective, that if I can write down the equation of the line that passes through radius which is (2,1,1) in the direction that is perpendicular to the plane (and thus can use the normal vector [a,2,1]) then I can just find the intersecting point. The equation of the line then must be $$r(t)= [2,1,1]+t[a,2,1]$$. From here we get the parametric equations $$x(t)=2+at$$ $$y(t)=1+2t$$ $$z(t)=1+t$$. Pugging in each equation into the plane equation gives me a quadratic in $a$. Now I am not sure if this is the right approach or not, or if I am even doing this right, but any clarification or confirmation if I'm doing the problem right will be much appreciated.