I need to find out the equation of circle having three points $$(a,0,0), (0,b,0), (0,0,c)$$.
I know that this represent when sphere is intersected by a plane. I can find the equation of plane using these planes by it's intercept form $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$, but how to find equation of sphere using 3 points only. Please suggest.
On
A completion on @ajotatxe answer:
The bisector plane of the first two points is given by $$(x-a)^2+y^2+z^2=x^2+(y-b)^2+z^2$$ That is, $$2ax-2by=a^2-b^2$$ Do the same for other bisector plane, for example $2ax-2cz=a^2-c^2$.
Do the same for last bisector plane, $2by-2cz=b^2-c^2$.
Although two of them is enough.
Find the intersection of selected two planes and $x/a+y/b+z/c=1$.
This is the center of the circle.
Then the radius can be computed from this formula: $$R = \frac{\alpha \beta \gamma}{2S}$$ Where $\alpha, \beta, \gamma$ are edges lenght and $S$ is the area of triangle that can be computed from Heron's formula: $$S^2=p(p-\alpha)(p-\beta)(p-\gamma)$$ Where $p$ is the half of perimeter: $$p=\frac{\alpha+\beta+\gamma}{2}$$ In this example edges lenghts are $\sqrt{a^2+b^2}, \sqrt{a^2+c^2}, \sqrt{b^2+c^2}$
Alternative method:
The bisector plane of the first two points is given by $$(x-a)^2+y^2+z^2=x^2+(y-b)^2+z^2$$ That is, $$2ax-2by=a^2-b^2$$ Do the same for other bisector plane, for example $2ax-2cy=a^2-c^2$.
Find the intersection of these two planes and $x/a+y/b+z/c=1$. This is the center of the circle.