Question

If $(2, 7, 3)$ is one end of a diameter of the sphere $x^2 + y^2 + z^2 - 6\,x -12\,y -2\,z + 20 = 0$, then the coordinates of the other end of the diameter are

• A. $(-2, 5, -1)$
• B. $(4, 5, 1)$
• C. $(2, -5, 1)$
• D. $(4, 5, -1)$

Answer 1

Answer: (D)

$(4, 5, -1)$

Answer 2

Given sphere is
$x^{2}+y^{2}+z^{2}-6 \,x-12\, y-2 z+20=0$
Centre $\equiv(3,6,1)$
Here, one end of diameter is $(2,7,3)$.
Let the other end of the diameter be $(x, y, z)$
Centre of the sphere will be the mid-point of the ends of diameter.
So, $(3,6,1)=\left(\frac{2+x}{2}, \frac{7+y}{2}, \frac{3+z}{2}\right)$
$\Rightarrow 2+x=6$
$\Rightarrow x=4$
$\Rightarrow 7+y=12$
$\Rightarrow y=5$
and $3+z=2$
$\Rightarrow z=-1$
Therefore, $(x, y, z) \equiv(4,5,-1)$