Question

If $(2,3,5)$ is one end of a diameter of the sphere $x^2+y^2+z^2-6x-12y-2z+20=0$, then the coordinates of the other end of the diameter are

Answer 1

Finding coordinates of other end of the diameter:

The equation of sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ where $u$, $v$, $w$ and $d$ are constants.

The center of sphere is $(-u,-v,-w)$.

The given equation of sphere is $x^2+y^2+z^2-6x-12y-2z+20=0$. So the center of sphere is $(3,6,1)$.

The coordinates of one end of diameter is $(2,3,5)$.

Let the coordinates of other end of diameter be $(a,b,c)$.

$\frac{2+a}{2}=3,\frac{3+b}{2}=6,\frac{5+c}{2}=1$

$a=4,b=9,c=-3$

Hence, the coordinates of other end of the diameter are $(4,9,-3)$.