Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.  
Plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
and assuming the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.  
I've not been able to find the value for $d$, putting the three points in the equation of sphere I got $3$ equations with $4$ variables.
 A: First, let's work out the plane of the circle.
The three triangle vertices are
$$\vec{a} = \left [ \begin{array}{c} a \\ 0 \\ 0 \end{array} \right ], \qquad \vec{b} = \left [ \begin{array}{c} 0 \\ b \\ 0 \end{array} \right ], \qquad \vec{c} = \left [ \begin{array}{c} 0 \\ 0 \\ c \end{array} \right ] \tag{1}\label{na1}$$
The equation for the plane is
$$\vec{p} \cdot \vec{n} - d = 0$$
where the plane normal $\vec{n}$ is 
$$\vec{n} = ( \vec{b} - \vec{a} ) \times ( \vec{c} - \vec{a} ) = \left [ \begin{array}{c} -a \\ b \\ 0 \end{array} \right ] \times \left [ \begin{array}{c} -a \\ 0 \\ c \end{array} \right ] = \left [ \begin{array}{c} b c \\ a c \\ a b \end{array} \right ]$$
and the distance from plane to origin $d$ is the same at all points on the plane, including the vertices,
$$\vec{a} \cdot \vec{n} = \vec{b} \cdot \vec{n} = \vec{c} \cdot \vec{n} = a b c$$
i.e., the full equation of the plane is
$$\vec{p} \cdot \vec{n} - a b c = 0, \qquad \vec{n} = \left [ \begin{array}{c} b c \\ a c \\ a b \end{array} \right ] \tag{2}\label{na2}$$
Now, one way to treat the circle is to treat it as a great circle of a sphere that passes through the points. The key point is that because it is a great circle, the center of the sphere must be on the same plane as the vertices.
This also means that we can use two coordinates, say $u$ and $v$, to describe the center of the circle (and sphere):
$$\vec{p} = \vec{a} + u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) \tag{3}\label{na3}$$
If the radius of the circle (and therefore the radius of the sphere) is $r$, then we have a system of equations,
$$\begin{cases}
\left \lVert \vec{p} - \vec{a} \right \rVert^2 = r^2 \\
\left \lVert \vec{p} - \vec{b} \right \rVert^2 = r^2 \\
\left \lVert \vec{p} - \vec{c} \right \rVert^2 = r^2
\end{cases}$$
i.e.
$$\begin{cases}
\left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) \right \rVert^2 - r^2 = 0 \\
\left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) + \vec{a} - \vec{b} \right \rVert^2 - r^2 = 0 \\
\left \lVert u ( \vec{b} - \vec{a} ) + v (\vec{c} - \vec{a}) + \vec{a} - \vec{c} \right \rVert^2 - r^2 = 0
\end{cases} \tag{4}\label{na4}$$
This system of equations has two solutions, one of which has a negative radius $r$; thus only one solution that makes sense:
$$\begin{cases}
u = \frac{a^2 b^2 + b^2 c^2}{2 (a^2 b^2 + a^2 c^2 + b^2 c^2)} \\
v = \frac{a^2 c^2 + b^2 c^2}{2 (a^2 b^2 + a^2 c^2 + b^2 c^2)} \\
r = \sqrt{\frac{(a^2 + b^2)(a^2 + c^2)(b^2 + c^2)}{4 (a^2 b^2 + a^2 c^2 + b^2 c^2 )}} \tag{5}\label{na5}\end{cases}$$
To find the center $\vec{p}$ in Cartesian coordinates, insert $u$ and $v$ and $\eqref{na1}$ to $\eqref{na3}$.
A: Given the points
$$
\cases{
A=(a,0,0)\\
B=(0,b,0)\\
C=(0,0,c)\\
p = (x,y,z)
}
$$
the sphere center $p_0$ is determined by the intersection of plane $\Pi$ with the line $L$ such that
$$
\cases{
\Pi\to (p-A)\cdot\vec n=0\\
\vec n = (A-B)\times(B-C)\\
L\to \left((p-\frac{A+B}{2})\cdot(A-B)=0\right)\cap \left((p-\frac{A+C}{2})\cdot(A-C)=0\right)\\
L\to p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)
}
$$
then
$$
p_0 = \frac{1}{2(a^2b^2+a^2c^2+b^2 c^2)}\left(a^3 \left(b^2+c^2\right),b^3 \left(a^2+c^2\right),c^3\left(a^2+b^2\right)\right)
$$
and
$$
r = \|p_0-A\| = \frac{1}{2} \sqrt{\frac{\left(a^2+b^2\right) \left(a^2+c^2\right) \left(b^2+c^2\right)}{a^2b^2+a^2c^2+b^2 c^2}}
$$
Attached a plot for $a=1,b=2,c=3$. In pink the plane $(p-A)\cdot \vec n=0$. In green the plane $(p-\frac{A+B}{2})\cdot(A-B)=0$ and in yellow the plane $(p-\frac{A+C}{2})\cdot(A-C)=0$. In red the line $p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)$

