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 PA²+PB²= k², where k is a constant.

Answer 1

If A and B be the points A(3, 4, 5) and B(–1, 3, –7), find the equation of the set of points P(x, y, z) such that PA² + PB² = k², where k is a constant.

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

\( PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2 \)

\( PB^2 = (x+1)^2 + (y-3)^2 + (z+7)^2 \)

Adding PA² and PB²:

\( (x-3)^2 + (y-4)^2 + (z-5)^2 + (x+1)^2 + (y-3)^2 + (z+7)^2 = k^2 \)

Expanding all terms:

\( (x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 - 10z + 25) \)

\( + (x^2 + 2x + 1) + (y^2 - 6y + 9) + (z^2 + 14z + 49) = k^2 \)

Combining like terms:

\( 2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2 \)

Dividing both sides by 2:

\( x^2 + y^2 + z^2 - 2x - 7y + 2z + \frac{109 - k^2}{2} = 0 \)

Required equation of the locus:
\( x^2 + y^2 + z^2 - 2x - 7y + 2z + \frac{109 - k^2}{2} = 0 \)