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Question

A variable plane passes through a fixed point (a,b,c) and cuts the axes in A,B and C respectively. The locus of the centre of the sphere OABC,O being the origin, is
  1. ax+by+cz=1
  2. xa+yb+zc=1
  3. ax+by+cz=2
  4. xa+yb+zc=2

A
ax+by+cz=1
B
ax+by+cz=2
C
xa+yb+zc=2
D
xa+yb+zc=1
Solution
Verified by Toppr

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,

axA+byB+czC=1

Let the centre of the sphere OABC be given by (X,Y,Z).
Then, the equation of the circle, with radius r:

(xX)2+(yY)2+(zZ)2=r2

Since O(0,0,0) lies on the sphere
(0X)2+(0Y)2+(0Z)2=r2
i.e

X2+Y2+Z2=r2

A, B, C also lie on the sphere. Hence,

(xAX)2+(0Y)2+(0Z)2=r2

(0X)2+(yBY)2+(0Z)2=r2

(0X)2+(0Y)2+(zCZ)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:

ax+by+cz=2

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