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 $P A^{2} + P B^{2} = k^{2}$ where k is a constant.

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Solution

Let $P (x, y, z)$ be any point

Then

$P A = \sqrt{(x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2}}$

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

$P B = \sqrt{(x + 1)^{2} + (y - 3)^{2} + (z + 7)^{2}}$

= $\sqrt{x^{2} + 1 + 2 x + y^{2} + 9 - 6 y + z^{2} + 49 - 14 z}$

Now $P A^{2} + P B^{2} = k^{2}$

$∴ {[\sqrt{x^{2} + 9 - 6 x + y^{2} + 16 - 8 y + z^{2} + 25 - 10 z}]}^{2} + {[\sqrt{x^{2} + 1 + 2 x + y^{2} + 9 - 6 y + z^{2} + 49 - 14 z}]}^{2} = k^{2}$

$∴ x^{2} + 9 - 6 x + y^{2} + 16 - 8 y + z^{2} + 25 - 10 z + x^{2} + 1 + 2 x + y^{2} + 9 - 6 y + z^{2} + 49 + 14 z = k^{2}$

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

$\Rightarrow 2 (x^{2} + y^{2} + z^{2} - 2 x - 7 y + 2 z)$

= $k^{2} - 109$

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

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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 PA2+PB2=k2 where k is a constant.

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 $P A^{2} + P B^{2} = k^{2}$ where k is a constant.