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.
Best Answer
The equation of a line through the center of sphere, perpendicular to the plane is $$ r(t)=(2,0,0)+t(a,2,1). $$ (I think that is what you meant in your attempt) The intersection with the plane is obtained by solving the system line-plane. Plugging into the equation of the plane you get $$ a(2+ta)+2(2t)+t=2\leftrightarrow t(a^2+5)=2-2a\leftrightarrow t=\frac{2a-2}{a^2+5} $$ So the point of intersection is $$ \left(2+\frac{2a^2-2a}{a^2+5},\frac{4a-4}{a^2+5},\frac{2a-2}{a^2+5}\right) $$ You need this point to be on the sphere, which gives you a condition on $a$ $$ [\frac{2a^2-2a}{a^2+5}]^2+[\frac{4a-4}{a^2+5}]^2+[\frac{2a-2}{a^2+5}]^2=4 $$ or $$ 4a^4-8a^3+24a^2-40a+20=4a^4+40a^2+100, $$ or $$ 8a^3+16a^2+40a+80=0\leftrightarrow a^3+2a^2+5a+10=0 $$ One solution is clearly $-2$. There no more real solutions. So the answer is $a=-2$. The fact that there is only one solution is telling us that the line $AB$ is tangent to the sphere, so only one plane in the family is tangent to the sphere. If the line were outside, you would have two possible planes and two possible values of $a$.