We are given two problems. Problem 43: An object's position $P$ changes so that its distance from $(1, 2, -3)$ is always twice its distance from $(1, 2, 3)$. We need to show that $P$ is on a sphere and find its center and radius. Problem 44: An object's position $P$ changes so that its distance from $(1, 2, -3)$ always equals its distance from $(2, 3, 2)$. We need to find the equation of the plane on which $P$ lies.

1. Problem Description

We are given two problems.

Problem 43: An object's position $P$ changes so that its distance from $(1, 2, -3)$ is always twice its distance from $(1, 2, 3)$. We need to show that $P$ is on a sphere and find its center and radius.

Problem 44: An object's position $P$ changes so that its distance from $(1, 2, -3)$ always equals its distance from $(2, 3, 2)$. We need to find the equation of the plane on which $P$ lies.

2. Solution Steps

Problem 43:

Let $P = (x, y, z)$. The distance from $P$ to $(1, 2, -3)$ is $\sqrt{(x-1)^2 + (y-2)^2 + (z+3)^2}$. The distance from $P$ to $(1, 2, 3)$ is $\sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2}$.

We are given that the distance from $P$ to $(1, 2, -3)$ is twice the distance from $P$ to $(1, 2, 3)$. Thus,

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

Squaring both sides, we get

$(x-1)^2 + (y-2)^2 + (z+3)^2 = 4[(x-1)^2 + (y-2)^2 + (z-3)^2]$

$(x-1)^2 + (y-2)^2 + (z+3)^2 = 4(x-1)^2 + 4(y-2)^2 + 4(z-3)^2$

$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 + 6z + 9 = 4(x^2 - 2x + 1) + 4(y^2 - 4y + 4) + 4(z^2 - 6z + 9)$

$x^2 - 2x + y^2 - 4y + z^2 + 6z + 14 = 4x^2 - 8x + 4y^2 - 16y + 4z^2 - 24z + 52$

$0 = 3x^2 - 6x + 3y^2 - 12y + 3z^2 - 30z + 38$

$3x^2 - 6x + 3y^2 - 12y + 3z^2 - 30z + 38 = 0$

$x^2 - 2x + y^2 - 4y + z^2 - 10z + \frac{38}{3} = 0$

Completing the square:

$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 10z + 25) = 1 + 4 + 25 - \frac{38}{3}$

$(x-1)^2 + (y-2)^2 + (z-5)^2 = 30 - \frac{38}{3} = \frac{90 - 38}{3} = \frac{52}{3}$

This is the equation of a sphere with center $(1, 2, 5)$ and radius $\sqrt{\frac{52}{3}} = 2\sqrt{\frac{13}{3}} = \frac{2\sqrt{39}}{3}$.

Problem 44:

Let $P = (x, y, z)$. The distance from $P$ to $(1, 2, -3)$ is $\sqrt{(x-1)^2 + (y-2)^2 + (z+3)^2}$. The distance from $P$ to $(2, 3, 2)$ is $\sqrt{(x-2)^2 + (y-3)^2 + (z-2)^2}$.

We are given that the distance from $P$ to $(1, 2, -3)$ equals the distance from $P$ to $(2, 3, 2)$. Thus,

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

Squaring both sides, we get

$(x-1)^2 + (y-2)^2 + (z+3)^2 = (x-2)^2 + (y-3)^2 + (z-2)^2$

$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 + 6z + 9 = x^2 - 4x + 4 + y^2 - 6y + 9 + z^2 - 4z + 4$

$-2x - 4y + 6z + 14 = -4x - 6y - 4z + 17$

$2x + 2y + 10z = 3$

$2x + 2y + 10z - 3 = 0$

This is the equation of a plane.

3. Final Answer

Problem 43:

The equation of the sphere is $(x-1)^2 + (y-2)^2 + (z-5)^2 = \frac{52}{3}$.

The center of the sphere is $(1, 2, 5)$.

The radius of the sphere is $\frac{2\sqrt{39}}{3}$.

Problem 44:

The equation of the plane is $2x + 2y + 10z - 3 = 0$.