For which value $a$ is the sphere $(x-2)^2+(y+1)^2+(z-2)^2\leq 10$ intersection with the plane x=a given by a cirle of area $6\pi$?

Question

Consider the sphere $(x-2)^2+(y+1)^2+(z-2)^2\leq 10$. Let its intersection with the plane x=a be given by a cirle of area $6\pi$.

What are the possible values of $a$?

Answer 1

The intersection of the sphere $(x-2)^2+(y+1)^2+(z-2)^2=10$ with the plane $x=a$ is given by:

$$(a-2)^2+(y+1)^2+(z-2)^2=10 \iff (y+1)^2+(z-2)^2=10-(a-2)^2$$

This is an equation of a circle in the $yz$ plane, with a squared radius of: $$R^2=10-(a-2)^2$$

We want this circle to be of area $6\pi$, so the formula for a circle's area yields:

$$6\pi=\pi R^2=\pi(10-(a-2)^2) \implies 10-(a-2)^2=6 \implies (a-2)^2=4 \implies a-2=\pm2$$ $$\implies \fbox{$a=0 \space\text{or} \space a=4$}$$