It 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 + (PB)2 = K2, where K is a constant.

A
$2 (x^{2} + y^{2} + z^{2}) + 4 x + 14 y + 4 z + 109 - K^{2} = 0$
B
$2 (x^{2} + y^{2} + z^{2}) - 4 x - 14 y + 4 z + 109 - K^{2} = 0$
C
$x^{2} + y^{2} + z^{2} + 4 x + 14 y + 4 z + 109 - K^{2} = 0$
D
None of the above

Solution:

Let the point P is (x, y, z).

Given (PA)² + (PB)² = K²

$\Rightarrow (x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2} + (x + 1)^{2} + (y - 3)^{2} + (z + 7)^{2} = K^{2}$

$\begin{array}{r} \Rightarrow x^{2} + 9 - 6 x + y^{2} + 16 - 8 y + z^{2} + 25 - 10 z + x^{2} + 1 + 2 x + y^{2} + 9 - 6 y + z^{2} + 49 + 14 z = K^{2} \end{array}$

$\begin{aligned} \Rightarrow 2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 - K^{2} = 0 \\ \Rightarrow 2 (x^{2} + y^{2} + z^{2}) - 4 x - 14 y + 4 z + 109 - K^{2} = 0 \end{aligned}$

which is the required equation.

It 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 + (PB)² = K², where K is a constant.

Solution:

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