# If A and B be the points (3, 4, 5) and (–1, 3, –7),

Question:

If A and B be the points $(3,4,5)$ and $(-1,3,-7)$ respectively, find the equation of the set of points P such that PA $^{2}+P B^{2}=k^{2}$, where $k$ is a constant.

Solution:

The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.

Let the coordinates of point P be (xyz).

On using distance formula, we obtain

$\mathrm{PA}^{2}=(x-3)^{2}+(y-4)^{2}+(z-5)^{2}$

$=x^{2}+9-6 x+y^{2}+16-8 y+z^{2}+25-10 z$

$=x^{2}-6 x+y^{2}-8 y+z^{2}-10 z+50$

$\mathrm{PB}^{2}=(x+1)^{2}+(y-3)^{2}+(z+7)^{2}$

$=x^{2}+2 x+y^{2}-6 y+z^{2}+14 z+59$

Now, if $\mathrm{PA}^{2}+\mathrm{PB}^{2}=k^{2}$, then

$\left(x^{2}-6 x+y^{2}-8 y+z^{2}-10 z+50\right)+\left(x^{2}+2 x+y^{2}-6 y+z^{2}+14 z+59\right)=k^{2}$

$\Rightarrow 2 x^{2}+2 y^{2}+2 z^{2}-4 x-14 y+4 z+109=k^{2}$

$\Rightarrow 2\left(x^{2}+y^{2}+z^{2}-2 x-7 y+2 z\right)=k^{2}-109$'

$\Rightarrow x^{2}+y^{2}+z^{2}-2 x-7 y+2 z=\frac{k^{2}-109}{2}$

Thus, the required equation is $x^{2}+y^{2}+z^{2}-2 x-7 y+2 z=\frac{k^{2}-109}{2}$.