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. 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 $3(x^2+y^2+z^2)-2(x+y+z)-1=0$
 A: 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.
A: Write the equation of a candidate sphere as
$$ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2. $$
Plugging in the points, we get
$$ (1 - x_0)^2 + y_0^2 + z_0^2 = r^2, \\
   x_0^2 + (1 - y_0)^2 + z_0^2 = r^2, \\
   x_0^2 + y_0^2 + (1 - z_0)^2 = r^2.
$$
Comparing the first two equations, we get
$$ (1 - x_0)^2 + y_0^2 = x_0^2 + (1 - y_0)^2 \implies 1 - 2x_0 + x_0^2 + y_0^2 = x_0^2 + 1 - 2y_0 + y_0^2 \implies x_0 = y_0. $$
Similarly we see that $x_0 = y_0 = z_0$ and so the center of the sphere must lie on the line $x = y = z$. Denoting the center by $(t,t,t)$ and returning to the first equation, we obtain
$$ 1 - 2t + 3t^2 = r^2 $$
so to minimize $r$ (or $r^2$), we need to minimize the quadratic $1 - 2t + 3t^2$. The minimum $\frac{2}{3}$ is attained at the vertex $t = \frac{1}{3}$ and so the equation is
$$ \left( x - \frac{1}{3} \right)^2 + \left( y - \frac{1}{3} \right)^2 + \left( z - \frac{1}{3} \right)^2 = \frac{2}{3} \iff \\
x^2 + y^2 + z^2 - \frac{2}{3}(x + y + z) + \frac{1}{3} = \frac{2}{3} \iff \\
3(x^2 + y^2 + z^2) - 2(x + y + z) - 1 = 0.$$
A: Maybe I'm naive. but the three points lie in a circle in a plane. That circle will be a cross section of the sphere. The radius of the sphere must be at least the radius of this cross section circle.  So the least possible radius would be exactly the radius of the circle (which would make the circle a great circle).
Find the center of the circle with these points and that will be the center of the sphere.
A: Let the equation of sphere is  $x^2+y^2+z^2+2ux+2vy+2wz+d=0$
It's given that the sphere passes through the points $(1,0,0),(0,1,0),(0,0,1)$
Therefore we get $u=v=w=-1/2(d+1)$
We know that $r^2=u^2+v^2+w^2-d$
                     Put the values of $u, v,w$:
$r^2=3/4((d+1/3)^2+8/4)$
We have to make $r$ as small as possible 
Therefore $r$  will be smallest if $d=-1/3$ hence put the values of u, v, W, and d in equation of sphere and solve it.
We get the required result
