Find the equation of the sphere with points P such that the distance from P to A(−2, 4, 4) is twice the distance from P to B(6, 3, −1).

Question

As you can tell from the title, I need to find the equation of the sphere with points $P$ such that $2|PB| = |PA|$. The coordinates for $A$ are $(-2, 4, 4)$ and the coordinates for $B$ are $(6, 3, -1)$. I've been stuck on this for two hours and can't seem to figure it out. I got as far as $x^2 + y^2 + z^2 - \frac{52}{3}x - \frac{16}{3}y+\frac{16}{3}z + \frac{148}{3} = 0$ but apparently this is wrong.

Thanks

Answer 1

Let an arbitrary point, P be (x,y,z). We have $A(-2,4,4)$ and $B(6, 3, -1)$ From the question, we know |PA| = 2|PB|. Therefore, $$ \sqrt{(x - (-2))^2 + (y-4)^2 + (z-4)^2} = 2\sqrt{(x - 6)^2 + (y-3)^2 + (z-(-1))^2}$$ $$(x +2)^2 + (y-4)^2 + (z-4)^2 = 4((x - 6)^2 + (y-3)^2 + (z+1)^2)$$ Using algebra, $$x^2+4x+4+y^2-8y+16+z^2-8z+16= 4(x^2-12x+36+y^2-6y+9+z^2+2z+1)$$ $$x^2+4x+4+y^2-8y+16+z^2-8z+16= 4x^2-48x+144+4y^2-24y+36+4z^2+8z+4$$ $$-3x^2+52x-140-3y^2+16y-20-3z^2-16z+12=0$$ $$x^2+y^2+z^2-(52/3)x-(16/3)y+(16/3)z+148/3=0$$ which is the form you were looking for. I am assuming there was some error somewhere in your calculations.

Answer 2

We connect $A$ to $B$ and extend the line to the far side of the sphere. The length $B$ to the far side of the sphere equals the length from $A$ to $B$ (putting this point twice as far from $A$ as it is from $B$). Divide the distance from $A$ to $B$ in the required $2:1$ ratio. Producing this figure.

The center and radius of our sphere.

$r^2 = \|\frac 23 (A-B)\|^2 = \|\frac 23 (-8, 1, 5)\|^2 = \frac {4}{9} (8^2+1^2 + 5^2) = 40\\ C = B - \frac 13 (A-B) = (6,3,-1) - \frac 13 (-8,1,5) = (\frac {26}{3},\frac {8}{3},-\frac {8}{3})$

$(x - \frac {26}{3})^2 + (y-\frac {8}{3})^2 + (z+\frac {8}{3})^2 = 40$

Answer 3

Find the equation of the sphere with points P such that the distance from P to A(−2, 4, 4) is twice the distance from P to B(6, 3, −1).

More generally, in n dimensions suppose that the distance from P to A is r times the distance from P to B.

I will show:

If $r=1$ then P is the plane through the midpoint of AB perpendicular to B-A.

If $r \ne 1$ then P is a sphere with center $C =\dfrac{r^2B-A}{r^2-1} $ and radius $\dfrac{r }{|r^2 - 1|}|B-A| $.

(This is well known, but I enjoyed working it out and others might find this useful.)

The square of the distance from P to A is $r^2$ the distance from P to B so

$\sum_{k=1}^n (p_k-a_k)^2 =r^2\sum_{k=1}^n (p_k-b_k)^2 $ or $\sum_{k=1}^n (p_k^2-2p_ka_k+a_k^2) =r^2\sum_{k=1}^n (p_k^2-2p_kb_k+b_k^2) $.

If $r=1$ then

$\begin{array}\\ 0 &=\sum_{k=1}^n (2p_k(b_k-a_k)+a_k^2-b_k^2)\\ &=2\sum_{k=1}^n (p_k(b_k-a_k)-\dfrac{b_k^2-a_k^2}{2})\\ &=2\sum_{k=1}^n \left(p_k(b_k-a_k)-\dfrac{(b_k-a_k)(b_k+a_k)}{2}\right)\\ &=2\sum_{k=1}^n \left(p_k(b_k-a_k)-\dfrac{b_k+a_k}{2}(b_k-a_k)\right)\\ &=2\sum_{k=1}^n \left((p_k-\dfrac{b_k+a_k}{2})(b_k-a_k)\right)\\ &=2\sum_{k=1}^n \left((p_k-c_k)(b_k-a_k)\right) \qquad c_k=\dfrac{b_k+a_k}{2}\\ \end{array} $

or, in vector terms, $(P-C)\cdot(B-A) =0 $ where $C=(A+B)/2 $ is the midpoint of $AB$, so P is the plane through the midpoint of AB perpendicular to B-A.

If $r^2\ne 1$ then

$\begin{array}\\ 0 &=\sum_{k=1}^n \left((r^2-1)p_k^2-2p_k(r^2b_k-a_k)+r^2b_k^2-a_k^2)\right)\\ &=(r^2-1)\sum_{k=1}^n \left(p_k^2-2p_k\dfrac{r^2b_k-a_k}{r^2-1}+\dfrac{r^2b_k^2-a_k^2}{r^2-1}\right)\\ &=(r^2-1)\sum_{k=1}^n \left(p_k^2-2p_kc_k+d_k\right) \quad c_k=\dfrac{r^2b_k-a_k}{r^2-1}, d_k=\dfrac{r^2b_k^2-a_k^2}{r^2-1}\\ \text{so}\\ 0 &=\sum_{k=1}^n \left(p_k^2-2p_kc_k+d_k\right)\\ &=\sum_{k=1}^n \left(p_k^2-2p_kc_k+c_k^2+d_k-c_k^2\right)\\ &=\sum_{k=1}^n \left((p_k^2-c_k)^2+d_k-c_k^2\right)\\ &=\sum_{k=1}^n (p_k-c_k)^2-\sum_{k=1}^n(c_k^2-d_k)\\ \end{array} $

Note that

$\begin{array}\\ c_k &=\dfrac{r^2b_k-a_k}{r^2-1}\\ &=\dfrac{r^2}{r^2-1}b_k-\dfrac{1}{r^2-1}a_k\\ \end{array} $

so that, if $C=(c_1, c_2, ..., c_n)$, then $C =\dfrac{r^2}{r^2-1}B-\dfrac{1}{r^2-1}A =\dfrac{r^2B-A}{r^2-1} $.

Also, since

$\begin{array}\\ c_k^2-d_k &=\dfrac{(r^2b_k-a_k)^2}{(r^2-1)^2}-\dfrac{r^2b_k^2-a_k^2}{r^2-1}\\ &=\dfrac{(r^2b_k-a_k)^2-(r^2b_k^2-a_k^2)(r^2-1)}{(r^2-1)^2}\\ &=\dfrac{r^4b_k^2-2r^2b_ka_k+a_k^2-(r^4b_k^2-r^2b_k^2-r^2a_k^2+a_k^2)}{(r^2-1)^2}\\ &=\dfrac{-2r^2b_ka_k-(-r^2b_k^2-r^2a_k^2)}{(r^2-1)^2}\\ &=\dfrac{r^2(b_k^2-2b_ka_k+a_k^2)}{(r^2-1)^2}\\ &=\dfrac{r^2 (b_k - a_k)^2}{(r^2 - 1)^2}\\ \end{array} $

$\begin{array}\\ \sum_{k=1}^n(c_k^2-d_k) &=\sum_{k=1}^n\dfrac{r^2 (b_k - a_k)^2}{(r^2 - 1)^2}\\ &=\dfrac{r^2 }{(r^2 - 1)^2}\sum_{k=1}^n (b_k - a_k)^2\\ &=\dfrac{r^2 }{(r^2 - 1)^2}|B-A|^2\\ \end{array} $

so the equation is

$\sum_{k=1}^n (p_k-c_k)^2 =\dfrac{r^2 }{(r^2 - 1)^2}|B-A|^2 $.

This is a sphere with center $C =\dfrac{r^2B-A}{r^2-1} $ and radius $\dfrac{r }{|r^2 - 1|}|B-A| $.