Show that the spheres, which cut two given spheres along a great circle, all pass through two fixed points.
let the given spheres be
x²+y²+z²+2$g_1$x+2$f_1$y+2$h_1$z+$d_1$=0 ...... (1) and
x²+y²+z²+2$g_2$x+2$f_2$y+2$h_2$z+$d_2$=0 ..... (2)
led the Sphene x²+y²+z²+2$g$x+2$f$y+2$h$z+$d$=0 .... (3) cuts the two spheres along a great circle.
Therefore, equation of plane of intersection of the spheres (1) and (3) is
2(f-f$_1$)+2(g-g$_1$)+2(h-h$_1$)+d-d$_1$ =0..... (4)
Perpendicular distance of the plane (4) from centre (-g$_1$,-f$_1$,-h$_1$) of the sphere (1) is equal to 0 Therefore 2(f-f$_1$)f$_1$+2(g-g$_1$)g$_1$+2(h-h$_1$)h$_1$-d+d$_1$ =0
or, (g$_1$)²+(f$_1$)²+(h$_1$)²-2gg$_1$-2ff$_1$-2hh$_1$+d-(r$_1$)²=0 Where r$_1$ = radius of sphere (1)
similar equation we can obtain for spheres (2) and (3)
I can't understand how to proceed further please help me
Let the first given sphere be centered at $C_1$ with a radius of $R_1$, and the second given sphere be centered at $C_2$ with a radius of $R_2$.
Now let a certain unknown sphere with a center, say, $C$, and radius $R$, cut both spheres in great circles, then
$ \| C C_1 \|^2 = R^2 - R_1^2 $
Similarly
$ \| C C_2 \|^2 = R^2 - R_2^2 $
If $C_1 = (x_1, y_1, z_1) , C_2 = (x_2, y_2, z_2) $ and $ C = (x, y, z) $, then these two equations become
$ (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = R^2 - R_1^2 $
$ (x - x_2)^2 + (y - y_2)^2 + (z - z_2)^2 = R^2 - R_2^2 $
Subtracting
$ - 2 x (x_1 - x_2) - 2 y (y_1 - y_2) - 2 z (z_1 - z_2) = A $
where $A = R_2^2 - R_1^2 - (x_1^2 + y_1^2 + z_1^2) + (x_2^2 + y_2^2 + z_2^2 )$
And this is an equation of a plane.
To simplify the analysis, we can assume WLOG that
$(x_1, y_1 , z_1) = (0,0,0) $
and
$(x_2, y_2, z_2) = (a, 0, 0) $
So now the equation of the plane is
$ 2 a x = R_2^2 - R_1^2 + a^2 $
so that
$ x = x_0 = \dfrac{1}{2} ( R_2^2 - R_1^2 + a^2 ) $
$ y = t$
$ z = s$
From above, we have
$ (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = R^2 - R_1^2 $
So,
$ x_0^2 + t^2 + s^2 = R^2 - R_1^2 $
which is a circle in the $yz$ plane with center $(x_0, 0, 0)$ and radius equal to $r = \sqrt{ R^2 - R_1^2 - x_0^2 } $
Thus all the spheres of a given $R$ will centered at
$ C = (x_0 , 0 , 0) + r \cos \theta (0, 1, 0) + r \sin \theta (0, 0, 1) $
From symmetry, the intersection of all these spheres is with the $x$ axis which is $y = z = 0$
Apply Pythagoras to any of these spheres gives the intersections points
$ X_1 = x_0 - \sqrt{R^2 - r^2} $ and $ X_2 = x_0 + \sqrt{R^2 - r^2} $
Substituting $r^2$ , this gives
$ X_1 = x_0 - \sqrt{ R_1^2 + x_0^2 } ,\ \ X_2 = x_0 + \sqrt{R_1^2 + x_0^2 } $
Recall that
$ x_0 = \dfrac{1}{2} ( R_2^2 - R_1^2 + a^2 ) $
So these intersections which are $(X_1, 0, 0)$, and $(X_2, 0, 0)$ are fixed and independent of $R$ and the location of the sphere center.