Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$ Find the center of the circle through the points $(-1,0,0),(0,2,0),(0,0,3).$

Let the circle passes through the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ and the plane $Ax+By+Cz+D=0$
So the equation of the circle is $x^2+y^2+z^2+2ux+2vy+2wz+d+\lambda(Ax+By+Cz+D)=0$
As it passes through $(-1,0,0)$ so
$1-2u+d+\lambda(-A+D)=0$
As it passes through $(0,2,0)$ so
$4+4v+d+\lambda(2B+D)=0$
As it passes through $(0,0,3)$ so
$9+6w+d+\lambda(3C+D)=0$
Here i am stuck.I cannot find the radius of the circle.Is my method correct or not?Is there simpler method possible?Please help.

follow this process 


*

*find the plane passing through the three points (as the 3 points are points on axis it is very simple)

*take normal vector to the plane 

*consider any two sides of the triangle made by the three points and take their mid-point

*now take cross product between side's vector and plane's normal vector

*this will give you direction of perpendicular to the side

*now take a line passing through the mid-point of that side

*do the same process for the second side

*the point these both points intersect is your center of the circle


we just intersected two perpendicular bisectors of the sides 
