Question

Show that the plane 2𝑥 + 𝑦 + 2𝑧 = 0 is a tangent plane to the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 − 2𝑥 + 2𝑦 − 2𝑧 + 2 = 0.

Answer 1

Now we know, If the plane touches the sphere, then the perpendicular distance from the center of the sphere to the plane should be equal to the radius of the given sphere.

We have $\displaystyle 𝑥^2 + 𝑦^2 + 𝑧^2 − 2𝑥 + 2𝑦 − 2𝑧 + 2 = 0$

On adding and subtracting terms and rearranging we get,

$\displaystyle (x-1)^2+(y+1)^2+(z-1)^2=1$ using method of completing the squares.

So, the center of the sphere is $(1,\ -1,\ 1)$ and the radius is $1$ since the general equation of a sphere is $(x-g)^2+(y-h)^2+(z-I)^2=r^2$ where r is the radius of the sphere.

Now, the distance Between $(1,\ -1,\ 1)$ and the plane $\displaystyle 2𝑥 + 𝑦 + 2𝑧 = 0$ is;

$\displaystyle \Rightarrow p=\frac{Ax+By+Cz+D}{\sqrt{A^2+B^2+C^2}}$ where $Ax+By+Cz+D$ is the equation of the plane with $x$ , $y$ and $z$ being the points through which the distance is to measured, which here is the center $(1,\ -1,\ 1)$.

$\displaystyle \Rightarrow p=\frac{2\cdot1+1\cdot(-1)+2\cdot 1+0}{\sqrt{2^2+1^2+2^2}}=1$

Therefore, the radius of the sphere and the distance '$p$' between the center of the sphere and the plane are equal.

Hence, the plane is tangential to the sphere.

Now we know that equation of the line joining the center and the plane’s point of contact can be written in the cartesian form as

$\displaystyle \frac{x−x_1}{A}=\frac{y−y_1}{B}=\frac{z−z_1}{C}=\lambda$

where $(x_1\ ,y_1,\ z_1)=(1,\ -1,\ 1)$ and $A$ , $B$ and $C$ being the coefficients of $x$ , $y$ and $z$ in the equation of the plane.

$\displaystyle \Rightarrow\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{2}=\lambda\Rightarrow x=2\lambda+1,\ y=\lambda-1,\ z=2\lambda+1$

Substituting these points in the equation of the plane we get:

$\displaystyle 0=2𝑥 + 𝑦 + 2𝑧=2(2\lambda+1)+(\lambda-1)+2(2\lambda+1)\\ \Rightarrow \lambda=-\frac{1}{3}\Rightarrow x=\frac{1}{3},\ y=-\frac{4}{3},\ z=\frac{1}{3}$

Hence the point of contact is;

$\displaystyle \left(\frac{1}{3},\ -\frac{4}{3},\ \frac{1}{3}\right)$