3danalytic geometry
Obtain the equation of the sphere which passes through the points $(1,0,0),(0,1,0),(0,0,1)$ and has its radius as small as possible.
Let the sphere passes through $(x_1,y_1,z_1)$
Then i obtained the equation of sphere as $(x^2+y^2+z^2)(x_1+y_1+z_1)-(x_1^2+y_1^2+z_1^2-1)(x+y+z)+x_1^2+y_1^2+z_1^2-x_1-y_1-z_1=0$
I am stuck here.I do not know how to minimize the radius.The answer given in my book is $3(x^2+y^2+z^2)-2(x+y+z)-1=0$
The equation of a sphere of radius $r$ centred at $(u,v,w)$ is
$(x - u)^2 + (y-v)^2 + (z-w)^2 = r^2$, or $x^2 + y^2 + z^2 + U x + V y + W z + C = 0$ where $C = u^2 + v^2 + w^2 - r^2$, $U = -2u$, $V=-2v$, $W=-2w$. If $A$ is the matrix you're taking the determinant of, to have this equation satisfied for the four given points and an arbitrary point $(x,y,z)$ on the sphere means
$$A \pmatrix{1\cr U\cr V\cr W\cr C\cr} = \pmatrix{0\cr 0\cr 0\cr 0\cr0\cr}$$
But for that to happen, $\det A$ must be $0$. Conversely, if $\det A = 0$, there exist $T,U,V,W,C$ not all $0$ for which $$A \pmatrix{T\cr U\cr V\cr W\cr C\cr} = \pmatrix{0\cr 0\cr 0\cr 0\cr0\cr}$$
If $T \ne 0$ we may divide by it and get an equation of a sphere going through the points. If $T = 0$ we have a degenerate case where the "sphere" is actually a plane.
Yes, this generalizes to any number of dimensions.
Suppose the points are $A,B,C$. Then intersect the equations of perpendicular bisectors of $AB$ and $BC$. This is the center of the desired circle. (with your notation $(p,q)$)
Now calculate the distance between $(p,q)$ and $A$. Now $r$ is also found.
Best Answer
If the center of the sphere is $(a,b,c)$, then the square of the distance from $(a,b,c)$ to each of the points $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ is the same. Solve these equations to get a simple relation among $a$, $b$, and $c$. Then minimize the distance from the center to any one of the points.