Find an equation for the largest sphere that is centered at $(2,1,-2)$ and intersects the sphere $x^2 +y^2+z^2=1$

Question

So, basically my process so far has been to find the distance between the two centers; that is, the distance between $(0,0,0)$ and $(2,1,-2)$. Using the distance formula, I get that the distance between the two centers is $3$. Then, since the the "unit" sphere has a radius of $1$, I add $1$ to $3$ so that the radius of the circle centered at $(2,1,-2)$ is $4$. I'm not sure if this is the correct way to go about this, since the problem specifically asks for intersection.

Answer 1

Yes, you are right. You just need to state the following two things, they are all based on triangle inequality: $|x-y|+|y-z| \ge |x-z|$ , here $x,y,z$ are 3 points in $\mathbb{R}^3$, and the equality holds if and only if $\exists t \in [0,1], y = tx + (1-t)z$, which means $y$ lies between $x,z$:

(1) For all $p \in S^2$,$|(2,1,-2)-p| \le 4$, here $S^2$ means the unit sphere, and that's what you've proved

(2)There exists a point $q \in S^2$ such that $|(2,1,-2)-q|=4$. In fact, you just need to draw a ray emanating from $(2,1,-2)$ to $(0,0,0)$, its (second) intersection with $S^2$ is the $q$ we want.(Due to the equality of triangle inequality holds)

Answer 2

The equation of the largest circle centered at $(2,1,-2)$ and intersecting $$x^2+y^2+z^2=1$$ is $$(x-2)^2+(y-1)^2+ (z+2)^2= 16.$$

Note that the radius of such a circle is the distance between the two cneters plus the radius of the smaller circle which is $3+1=4.$