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.

Given: The points A (3, 4, 5) and B (-1, 3, -7)

⇒ x1 = 3, y1 = 4, z1 = 5; x2 = -1, y2 = 3, z2 = -7;


PA2 + PB2 = k2 ……….(i)


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


Now, by Distance Formula, we know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by .


So,


And


Now, substituting these values in (i), we have


[(3 - x)2 + (4 - y)2 + (5 - z)2] + [(-1 - x)2 + (3 - y)2 + (-7 - z)2] = k2


⇒ [(9 + x2 – 6x) + (16 + y2 – 8y) + (25 + z2 – 10z)] + [(1 + x2 + 2x) + (9 + y2 – 6y) + (49 + z2 + 14z)] = k2


⇒ 9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z + 1 + x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z = k2


⇒ 2x2 + 2y2 + 2z2 - 4x - 14y + 4z + 109 = k2


⇒ 2x2 + 2y2 + 2z2 - 4x - 14y + 4z = k2 – 109


⇒ 2 (x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109



Hence, the required equation is .


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