Constant Distance Ratio Sphere

Let's take $\sqrt{(x - 2)^2 + (y - 1)^2 + (z - 3)^2} = 2\sqrt{(x - 10)^2 + (y - 10)^2 + (z - 15)^2}$ ;

or $(x - 2)^2 + (y - 1)^2 + (z - 3)^2 = 4[(x - 10)^2 + (y - 10)^2 + (z - 15)^2];$

or $(x^2 - 4x + 4) + (y^2 - 2y + 1) + (z^2 - 6z + 9) = 4[(x^2 - 20x + 100) + (y^2 - 20y + 100) + (z^2 - 30z + 225);$

or $14 = 3x^2 + 3y^2 + 3z^2 -(80-4)x - (80-2)y - (120-6)z + 1700;$

or $14 - 1700 = (3x^2 - 76x) + (3y^2 - 78y) + (3z^2 - 114z);$

or $-1686 = 3(x^2 - \frac{76}{3}x) + 3(y^2 - 26y) + 3(z^2 - 38z);$

or $-\frac{1686}{3} = (x^2 - \frac{38}{3}x + (\frac{38}{3})^2 - (\frac{38}{3})^2) + (y^2 - 13y + 169 - 169) + (z^2 - 19z + 361 - 361);$

or $-\frac{1686}{3} + (\frac{38}{3})^2 + 169 + 361 = (x - \frac{38}{3})^2 + (y - 13)^2 + (z - 19)^2;$

or $\frac{1156}{9} = (x - \frac{38}{3})^2 + (y - 13)^2 + (z - 19)^2;$

Thus, the locus is a sphere centered at $(\frac{38}{3}, 13, 19)$ and with radius $\boxed{\sqrt{\frac{1156}{9}} \approx 11.33 }.$

Tom Engelsman offers a proof of why it's a sphere. But if we assume it must be a sphere, then if $P$ is where it passes through the line between $A$ and $B$ , and $Q$ is where it passes through the line outside of the segment $AB = 17$ , then we know that $AP = \dfrac{17}{3}$ , and $AQ = 17$ .
The radius is $\dfrac{1}{2}$ of $AP + AQ = \dfrac{34}{3} = 11.333...$

Obviously $1:2 = (\dfrac{17}{3}):(\dfrac{34}{3}) = (AP)(AB - AP)$ and $1:2 = (17):(34) = (AQ):(AB + AQ)$

We are given that,

$PA = k PB, k > 0$

So, in terms of the vectors $P$ , $A$ , and $B$ , this becomes,

$(P - A)^T (P - A) = k^2 (P - B)^T (P - B)$

Expanding,

$P^T P - 2 A^T P + A^T A = k^2 ( P^T P - 2 B^T P + B^T B)$

Rearranging,

$(k^2 - 1) P^T P + 2 (A - k^2 B)^T P + k^2 B^T B - A^T A = 0$

Define $P_0 = -\dfrac{1}{k^2 - 1} (A - k^2 B)$ , then

$(k^2 - 1) (P - P_0)^T (P - P_0) = A^T A - k^2 B^T B + (k^2 - 1) P_0^T P_0$

Now,

$P_0^T P_0 = \dfrac{1}{(k^2 - 1)^2} (A^T A + k^4 B^T B - 2 k^2 A^T B )$

Therefore,

$\begin{aligned} (k^2 - 1) & (P - P_0)^T (P - P_0) = A^T A - k^2 B^T B + \dfrac{1}{k^2 - 1} (At A + k^4 B^T B - 2k^2 A^T B) \\ &=\dfrac{1}{k^2-1} \{ (k^2 - 1) A^T A - k^2 (k^2 - 1) B^TB + A^T A + k^4 B^T B - 2 k^2 A^T B \} \\ &= \dfrac{1}{k^2 - 1} ( k^2 A^T A + k^2 B^T B - 2 k^2 A^T B ) \\ &= \dfrac{k^2}{k^2 - 1} (A - B)^T(A - B) \end{aligned}$ Hence,

$(P - P_0)^T (P - P_0) = (\dfrac{k}{k^2 - 1})^2 (A - B)^T (A - B) = r^2$

This is an equation of a sphere with center $P_0$ and radius $r$ , where

$P_0 = \dfrac{1}{k^2 -1 }(k^2 B - A)$

and,

$r = | \dfrac{k} { k^2 - 1} | | A - B |$

Substituting $k = 2$ , and $A = (2, 1, 3)$ and $B = (10, 10, 15)$ , we obtain

$r = | \dfrac{2 }{2^2 - 1}| | (2, 1, 3) - (10, 10, 15) | = \dfrac{2}{3} | (-8, -9, -12) | = \dfrac{2}{3} (17) = \dfrac{34}{3} = 11.33$ .

By symmetry, the center of the sphere has to be on the line connecting A and B.

On this line we can identify two points that have PA =2PB:

• one is between A and B, where PA = 2/3 AB and PB = 1/3 AB
• one is 'past B', as seen from A, where PA = 2AB and PB = AB

So these two points are 4/3 AB apart, so our radius must be $r=\frac{2}{3}AB=\frac{2}{3}\sqrt{8^2+9^2+12^2}=\frac{34}{3}=11.333....$

This locus of points at distances 1:2 from two fixed points, is a known way of defining a circle as well. The radius of the circle is 2/3 the distance between the two fixed points. So the radius will be (Distance 17) x 2 / 3 = 11.33

(Sorry, IMGUR is temporarily down, i will upload the image when it allows me to log in)

Here is a proof of this result:

Consider a circle of radius R and center C. Points A & B lie on a radial line at a distance R/2 and R from the point Q on the circle.

Now take a point P at an arbitrary distance L from A and 2L from B. Let CP = x.

Let's check if point P lies on the circle, using Stewart's theorem in the triangle shown:

$\frac{3R}{2}\times2R\times\frac{R}{2} + 2R\times L^2 = X^2 \times \frac{3R}{2} + 4L^2 \frac{R}{2}$

which gives X = R i.e. for any arbitrary distance L, point P always lies at a distance R from the center C, thus tracing the circle.