So, I have the equation of a sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ and a point, $P(2, 3, -6)$. I am supposed to find the point on the sphere farthest away from $P$.
I tried to figure out how to use parametric equations to represent the sphere and maximize that using the distance formula. However, they are multiple variables in this equation, and this led me to nowhere.
On
The center of the sphere is $C(5,-1,7)$. The line passing through $C$ and $P(2,3,-6)$ has parametric equations $$ \begin{cases} x=5+(2-5)t&x=5-3t\\ y=-1+(3-(-1))t&y=-1+4t\\ z=7+(-6-7)t&z=7-13t\\ \end{cases} $$ Substitute in the equation of the sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ $$(5-3t-5)^2+(-1+4t+1)^2+(7-13t-7)^2=69$$ $$194 t^2 = 69\to t=\pm\sqrt{\frac{69}{194}}$$ the point on the sphere farthest away from $P$ is: $$\left(3 \sqrt{\frac{69}{194}}+5,-2 \sqrt{\frac{138}{97}}-1,13 \sqrt{\frac{69}{194}}+7 \right)$$
The line connecting $P(2,3,-6)$ and the farthest point $M(a,b,c)$ is parallel to $M$’s normal surface vector $(a-5,b+1,c-7)$, i.e. $$\frac{a-2}{a-5}=\frac{b-3}{b+1}=\frac{c+6}{c-7}$$
Substitute into $(a-5)^2+(b+1)^2+(c-7)^2=69$ to obtain the point $$M\left(5+3\sqrt{\frac{69}{194}}, -1-4\sqrt{\frac{69}{194}}, 7+13\sqrt{\frac{69}{194}}\right) $$