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Q.
The equation of a sphere is $x^2 + y^2 + z^2 -10z = 0$. If one end point of a diameter of the sphere is (- 3, - 4, 5), what is the other end point ?
Three Dimensional Geometry
Solution:
The equation of the given sphere is
$x^2 + y^2 + z^2 - 10z = 0.$
$ \therefore \quad$ Its centre is $\left(0, 0, 5\right)$.
Coordinates of one end point of a diameter of the sphere is given as $\left( - 3, - 4, 5\right)$.
Let Coordinates of another end point of this diameter $\left(x_{1}, y_{1}, z_{1}\right)$
$\therefore \quad \frac{-3+x_{1}}{2} = 0\quad\Rightarrow x_{1} = 3$
$\frac{-4+y_{1}}{2} = 0\quad\Rightarrow y_{1} = 4$
and $\frac{5+z_{1}}{2} = 5 \Rightarrow z_{1} = 5$
$\therefore \quad$ Required coordinates are $\left(3, 4, 5\right).$