1
$\begingroup$

Let $P$ is a variable point moves such that $OP = 5$ and $AP = 13$. If $A(6, 12,4)$, $O$ is origin and then we have to find the length of path traced by $P$.

My attempt

$$(x - 6)^2 + (y - 12)^2 + (z - 4)^2 = 13^2$$

$$\because x^2+y^2+z^2=25$$

$$\therefore12x + 24y + 8z- 196 = 25 - 169$$

$$\Rightarrow12x + 24y + 8z = 52 $$

$$\Rightarrow3x+6y+4z=13~~~~ $$

But now how to proceed.

  • 0
    You have a point, that has a constant distance from two other points. If you care about just one point, $P$ would trace a sphere. So at points, where both apply, you must be in the crossing of thwo speres, a circle. – 2017-01-12

1 Answers 1

1

As you state in your attempt, point $P$ lies on the intersection of two spheres $S_A$ and $S_O$: $$S_A: (x - 6)^2 + (y - 12)^2 + (z - 4)^2 = 13^2,$$ $$S_O:x^2+y^2+z^2=5^2.$$ That intersection of two spheres is a circle. The equation you obtained, $3x+6y+2z=13$, is simply the plane where that circle lies.

So now you can compute the distance from the origin to the plane and use the radius of $S_O$ and Pythagorean Theorem to compute the radius of the circle.

Finally, knowing that the circle center lies on the line between $O$ and $P$ and the distance from $O$, you can also compute its coordinates.