Maxmum distance from a point to a sphere

Question

Which point of the sphere $x^{2} + y^{2} + z^{2} = 1$ is at the maximum distance from the point $\left(\, 2,1,3\, \right)$ ?.

I know that the point is outside the sphere. Should we proceed with maxima and minima concept ?.

Answer 1

You draw a line from $(2,1,3)$ to the center of the sphere $(0,0,0)$. The line meets the sphere at two points. One is the minimum distance and the other is the maximum distance. As the distance from $(0,0,0)$ to any point 0n the sphere is $1$, the maximum distance is

$$\sqrt {2^2+1^2+3^2} + 1$$

Answer 2

WLOG any point on the sphere $(\cos u,\sin u\cos t,\sin u\sin t)$

$$D^2=(\cos u-2)^2+(\sin u\cos t-1)^2+(\sin u\sin t-3)^2$$

$$2^2+1^2+3^2+\cos^2u+\sin^2u(\cos^2t+\sin^2t)-4\cos u-2\sin u(\cos t+3\sin t)$$

$$=2^2+1^2+3^2+\cos^2u+\sin^2u-4\cos u-2\sin u(\cos t+3\sin t)$$

$\cos t+3\sin t=\sqrt{1^2+3^2}\cos\left(t-\arccos\dfrac1{\sqrt{1^2+3^2}}\right)$

$\implies-\sqrt{10}\le\cos t+3\sin t\le\sqrt{10}$

$\implies-2\sqrt{10}|\sin u|\le-2\sin u(\cos t+3\sin t)\le2\sqrt{10}|\sin u|$

$$D^2-(2^2+1^2+3^2+1)\le-4\cos u+2\sqrt{10}|\sin u|\le\sqrt{4^2+(2\sqrt{10})^2}=2\sqrt{14}$$