Let the points on the coordinate axes be
A(xA,0,0)
B(0,yB,0)
C(0,0,zC).
Then the equation of the plane which passes through these points, given the x-, y- and z-intercepts, is given as:
⇒xxA+yyB+zzC=1
The fixed point (a,b,c) lies on the plane. Hence,
Let the centre of the sphere OABC be given by (X,Y,Z).
Then, the equation of the circle, with radius r:
⇒(x−X)2+(y−Y)2+(z−Z)2=r2
Since O(0,0,0) lies on the sphere
⇒(0−X)2+(0−Y)2+(0−Z)2=r2
i.e
⇒X2+Y2+Z2=r2
A, B, C also lie on the sphere. Hence,
⇒(xA−X)2+(0−Y)2+(0−Z)2=r2
⇒(0−X)2+(yB−Y)2+(0−Z)2=r2
⇒(0−X)2+(0−Y)2+(zC−Z)2=r2
Comparing the 4 equations above:
⇒xA=2X
⇒yB=2Y
⇒zC=2Z
It was already shown that
⇒axA+byB+czC=1
∴a2X+b2Y+c2Z=1
⇒aX+bY+cZ=2
Hence, the locus of the centres of the sphere (X,Y,Z) is given by: